INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(139, 14239, '# Grade 10 Optional Mathematics: Polynomials Study Guide\r\n## Unit: Algebra | Chapter: Polynomials | Complete Solutions\r\n\r\n---\r\n\r\n## 🎯 Chapter Introduction\r\n\r\nThis guide provides a comprehensive overview of the **Polynomials** chapter for Grade 10 Optional Mathematics. It includes clear, step-by-step solutions to a variety of problems involving the **Remainder Theorem**, **Factor Theorem**, and **Synthetic Division**. 🧠\r\n\r\n### Key Learning Areas\r\n* **Remainder Theorem:** Finding the remainder when a polynomial is divided by a linear expression.\r\n* **Factor Theorem:** Determining if a linear expression is a factor of a polynomial.\r\n* **Synthetic Division:** A shorthand method for polynomial division by a linear factor.\r\n* **Solving Polynomial Equations:** Finding the roots of cubic polynomial equations.\r\n\r\n---\r\n\r\n## 📚 Exercise Solutions\r\n\r\n### Short Questions\r\n\r\n***\r\n\r\n#### Question 1: Finding a Value Using Factor Theorem\r\n**If $x - 2$ is a factor of $2x^3 - 6x^2 - 5m - 2$, find the value of $m$.**\r\n\r\n**Solution:**\r\nLet the polynomial be $P(x) = 2x^3 - 6x^2 - 5m - 2$.\r\nThe divisor is $d(x) = x - 2$. Comparing this to $x - a$, we get $a = 2$.\r\n\r\nBy the Factor Theorem, if $(x - a)$ is a factor of $P(x)$, then $P(a) = 0$.\r\n$$\r\n\\begin{aligned}\r\nP(2) &= 0 \\\\\r\n2(2)^3 - 6(2)^2 - 5m - 2 &= 0 \\\\\r\n2(8) - 6(4) - 5m - 2 &= 0 \\\\\r\n16 - 24 - 5m - 2 &= 0 \\\\\r\n-10 - 5m &= 0 \\\\\r\n-10 &= 5m \\\\\r\nm &= -2\r\n\\end{aligned}\r\n$$\r\nThus, **$m = -2$**.\r\n\r\n***\r\n\r\n#### Question 2: Finding a Value Using Factor Theorem\r\n**If $(x + 2)$ is a factor of the polynomial $2x^3 + 3x^2 - mx + 4$, find the value of $m$.**\r\n\r\n**Solution:**\r\nLet $P(x) = 2x^3 + 3x^2 - mx + 4$.\r\nThe divisor is $d(x) = x + 2$. Comparing this to $x - a$, we get $a = -2$.\r\n\r\nBy the Factor Theorem, $P(-2) = 0$.\r\n$$\r\n\\begin{aligned}\r\n2(-2)^3 + 3(-2)^2 - m(-2) + 4 &= 0 \\\\\r\n2(-8) + 3(4) + 2m + 4 &= 0 \\\\\r\n-16 + 12 + 2m + 4 &= 0 \\\\\r\n0 + 2m &= 0 \\\\\r\n2m &= 0 \\\\\r\nm &= 0\r\n\\end{aligned}\r\n$$\r\nThus, **$m = 0$**.\r\n\r\n***\r\n\r\n#### Question 3: Finding a Value Using Remainder Theorem\r\n**If $2x^3 - 4x^2 + kx + 10$ is divided by $(x + 2)$, the remainder is 4. Find the value of $k$.**\r\n\r\n**Solution:**\r\nLet $P(x) = 2x^3 - 4x^2 + kx + 10$.\r\nThe divisor is $d(x) = x + 2$, so $a = -2$. The remainder is $R = 4$.\r\n\r\nBy the Remainder Theorem, $P(a) = R$.\r\n$$\r\n\\begin{aligned}\r\nP(-2) &= 4 \\\\\r\n2(-2)^3 - 4(-2)^2 + k(-2) + 10 &= 4 \\\\\r\n2(-8) - 4(4) - 2k + 10 &= 4 \\\\\r\n-16 - 16 - 2k + 10 &= 4 \\\\\r\n-22 - 2k &= 4 \\\\\r\n-2k &= 26 \\\\\r\nk &= -13\r\n\\end{aligned}\r\n$$\r\nThus, **$k = -13$**.\r\n\r\n***\r\n\r\n#### Question 4: Finding a Value Using Factor Theorem\r\n**If $x - 3$ is a factor of $2x^3 - x^2 + 10x - k$, find the value of $k$.**\r\n\r\n**Solution:**\r\nLet $P(x) = 2x^3 - x^2 + 10x - k$.\r\nThe divisor is $d(x) = x - 3$, so $a = 3$.\r\n\r\nBy the Factor Theorem, $P(3) = 0$.\r\n$$\r\n\\begin{aligned}\r\n2(3)^3 - (3)^2 + 10(3) - k &= 0 \\\\\r\n2(27) - 9 + 30 - k &= 0 \\\\\r\n54 - 9 + 30 - k &= 0 \\\\\r\n75 - k &= 0 \\\\\r\nk &= 75\r\n\\end{aligned}\r\n$$\r\nThus, **$k = 75$**.\r\n\r\n***\r\n\r\n#### Question 5: Finding a Value Using Factor Theorem\r\n**If $x - 5$ is a factor of $x^3 + px^2 + 4x + 5$, find the value of $p$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 + px^2 + 4x + 5$.\r\nThe divisor is $d(x) = x - 5$, so $a = 5$.\r\n\r\nBy the Factor Theorem, $P(5) = 0$.\r\n$$\r\n\\begin{aligned}\r\n(5)^3 + p(5)^2 + 4(5) + 5 &= 0 \\\\\r\n125 + 25p + 20 + 5 &= 0 \\\\\r\n150 + 25p &= 0 \\\\\r\n25p &= -150 \\\\\r\np &= -6\r\n\\end{aligned}\r\n$$\r\nThus, **$p = -6$**.\r\n\r\n***\r\n\r\n#### Question 6: Finding the Remainder\r\n**Using the Remainder Theorem, find the remainder when $8x^3 - 4x^2 + 2x - 5$ is divided by $2x - 1$.**\r\n\r\n**Solution:**\r\nLet $P(x) = 8x^3 - 4x^2 + 2x - 5$.\r\nFor the divisor $2x - 1$, we set it to zero to find the value to substitute: $2x - 1 = 0 \\implies x = \\frac{1}{2}$.\r\n\r\nBy the Remainder Theorem, the remainder is $P(\\frac{1}{2})$.\r\n$$\r\n\\begin{aligned}\r\nP\\left(\\frac{1}{2}\\right) &= 8\\left(\\frac{1}{2}\\right)^3 - 4\\left(\\frac{1}{2}\\right)^2 + 2\\left(\\frac{1}{2}\\right) - 5 \\\\\r\n&= 8\\left(\\frac{1}{8}\\right) - 4\\left(\\frac{1}{4}\\right) + 1 - 5 \\\\\r\n&= 1 - 1 + 1 - 5 \\\\\r\n&= -4\r\n\\end{aligned}\r\n$$\r\nThe remainder is **-4**.\r\n\r\n***\r\n\r\n#### Question 7: Using the Factor Theorem\r\n**If $x - 5$ is a factor of $2x^3 - 7x + (p - 12)$, find the value of p.**\r\n\r\n**Solution:**\r\nLet $P(x) = 2x^3 - 7x + p - 12$.\r\nSince $x-5$ is a factor, $a=5$ and $P(5)=0$.\r\n$$\r\n\\begin{aligned}\r\nP(5) &= 0 \\\\\r\n2(5)^3 - 7(5) + p - 12 &= 0 \\\\\r\n2(125) - 35 + p - 12 &= 0 \\\\\r\n250 - 35 + p - 12 &= 0 \\\\\r\n203 + p &= 0 \\\\\r\np &= -203\r\n\\end{aligned}\r\n$$\r\n**Note:** There appears to be a typo in the source material\'s calculation. The correct answer based on the polynomial is $p=-203$.\r\n\r\n***\r\n\r\n#### Question 8: Using Synthetic Division\r\n**If $x^3 - 21x - 20 = (x + 1) \\cdot q(x)$, find $q(x)$ by using synthetic division.**\r\n\r\n**Solution:**\r\nThe polynomial is $P(x) = x^3 + 0x^2 - 21x - 20$.\r\nThe divisor is $x+1$, so we use $a = -1$ for synthetic division.\r\n\r\n-1 | 1 0 -21 -20 | -1 1 20\r\n 1  -1   -20     0 \r\n\r\nThe quotient\'s coefficients are 1, -1, and -20. The remainder is 0.\r\nThus, **$q(x) = x^2 - x - 20$**.\r\n\r\n***\r\n\r\n#### Question 9: Using Synthetic Division\r\n**If $x^3 - 19x - 30 = (x + 2) \\cdot q(x)$, find $q(x)$ by using synthetic division.**\r\n\r\n**Solution:**\r\nThe polynomial is $P(x) = x^3 + 0x^2 - 19x - 30$.\r\nThe divisor is $x+2$, so we use $a = -2$.\r\n\r\n-2 | 1 0 -19 -30 | -2 4 30\r\n 1  -2   -15     0 \r\n\r\nThe quotient\'s coefficients are 1, -2, and -15. The remainder is 0.\r\nThus, **$q(x) = x^2 - 2x - 15$**.\r\n\r\n***\r\n\r\n#### Question 10: State and Apply the Factor Theorem\r\n**State the Factor Theorem. Use it to determine whether $x + 3$ is a factor of $x^3 - 8x + 3$.**\r\n\r\n**Solution:**\r\n**Factor Theorem:** A polynomial $P(x)$ has a factor $(x - a)$ if and only if $P(a) = 0$.\r\n\r\nLet $P(x) = x^3 - 8x + 3$. To check if $(x+3)$ is a factor, we test $P(-3)$.\r\n$$\r\n\\begin{aligned}\r\nP(-3) &= (-3)^3 - 8(-3) + 3 \\\\\r\n&= -27 + 24 + 3 \\\\\r\n&= 0\r\n\\end{aligned}\r\n$$\r\nSince $P(-3) = 0$, **yes, $(x+3)$ is a factor** of the polynomial.\r\n\r\n***\r\n\r\n#### Question 11: State and Apply the Remainder Theorem\r\n**State the Remainder Theorem. When $x^3 + 5x^2 - 6x - 16$ is divided by $(x + 2)$, find the remainder.**\r\n\r\n**Solution:**\r\n**Remainder Theorem:** If a polynomial $P(x)$ is divided by $(x - a)$, the remainder is $P(a)$.\r\n\r\nLet $P(x) = x^3 + 5x^2 - 6x - 16$. The divisor is $(x+2)$, so $a = -2$. The remainder is $P(-2)$.\r\n$$\r\n\\begin{aligned}\r\nP(-2) &= (-2)^3 + 5(-2)^2 - 6(-2) - 16 \\\\\r\n&= -8 + 5(4) + 12 - 16 \\\\\r\n&= -8 + 20 + 12 - 16 \\\\\r\n&= 8\r\n\\end{aligned}\r\n$$\r\nThe remainder is **8**.\r\n\r\n***\r\n\r\n#### Question 12: Finding a Value Using Remainder Theorem\r\n**If the polynomial $x^3 + 6x^2 + kx + 10$ is divided by $(x + 2)$, the remainder is 4. Find the value of $k$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 + 6x^2 + kx + 10$. The divisor is $(x+2)$, so $a=-2$. The remainder is $R=4$.\r\n\r\nBy the Remainder Theorem, $P(-2) = 4$.\r\n$$\r\n\\begin{aligned}\r\n(-2)^3 + 6(-2)^2 + k(-2) + 10 &= 4 \\\\\r\n-8 + 6(4) - 2k + 10 &= 4 \\\\\r\n-8 + 24 - 2k + 10 &= 4 \\\\\r\n26 - 2k &= 4 \\\\\r\n22 &= 2k \\\\\r\nk &= 11\r\n\\end{aligned}\r\n$$\r\nThus, **$k = 11$**.\r\n\r\n***\r\n\r\n#### Question 13: Finding a Value Using Factor Theorem\r\n**If $(x + 2)$ is a factor of $x^3 - 19x - p$, find the value of $p$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 - 19x - p$. Since $(x+2)$ is a factor, $a=-2$ and $P(-2)=0$.\r\n$$\r\n\\begin{aligned}\r\nP(-2) &= 0 \\\\\r\n(-2)^3 - 19(-2) - p &= 0 \\\\\r\n-8 + 38 - p &= 0 \\\\\r\n30 - p &= 0 \\\\\r\np &= 30\r\n\\end{aligned}\r\n$$\r\nThus, **$p = 30$**.\r\n\r\n***\r\n\r\n#### Question 14: Finding a Value Using Factor Theorem\r\n**If $(x - m)$ is a factor of $x^3 - mx^2 - 4x + m + 12$, calculate the value of $m$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 - mx^2 - 4x + m + 12$. Since $(x-m)$ is a factor, $a=m$ and $P(m)=0$.\r\n$$\r\n\\begin{aligned}\r\nP(m) &= 0 \\\\\r\n(m)^3 - m(m)^2 - 4(m) + m + 12 &= 0 \\\\\r\nm^3 - m^3 - 4m + m + 12 &= 0 \\\\\r\n-3m + 12 &= 0 \\\\\r\n12 &= 3m \\\\\r\nm &= 4\r\n\\end{aligned}\r\n$$\r\nThus, **$m = 4$**.\r\n\r\n***\r\n\r\n#### Question 15: Finding a Value Using Factor Theorem\r\n**If $x - 2$ is a factor of $x^3 - kx^2 + x - 6$, find the value of $k$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 - kx^2 + x - 6$. Since $(x-2)$ is a factor, $a=2$ and $P(2)=0$.\r\n$$\r\n\\begin{aligned}\r\nP(2) &= 0 \\\\\r\n(2)^3 - k(2)^2 + 2 - 6 &= 0 \\\\\r\n8 - 4k - 4 &= 0 \\\\\r\n4 - 4k &= 0 \\\\\r\n4 &= 4k \\\\\r\nk &= 1\r\n\\end{aligned}\r\n$$\r\nThus, **$k=1$**.\r\n\r\n***\r\n\r\n#### Question 16: Using the Factor Theorem\r\n**Use the Factor Theorem to determine whether $x + 4$ is a factor of the polynomial $x^3 - 15x + 4$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 - 15x + 4$. To check if $(x+4)$ is a factor, we test if $P(-4) = 0$.\r\n$$\r\n\\begin{aligned}\r\nP(-4) &= (-4)^3 - 15(-4) + 4 \\\\\r\n&= -64 + 60 + 4 \\\\\r\n&= 0\r\n\\end{aligned}\r\n$$\r\nSince $P(-4) = 0$, **yes, $(x+4)$ is a factor**.\r\n\r\n***\r\n\r\n#### Question 17: Using the Remainder Theorem\r\n**Find the remainder when the polynomial $x^6 - 1$ is divided by $x + 1$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^6 - 1$. The divisor is $(x+1)$, so $a=-1$. The remainder is $P(-1)$.\r\n$$\r\n\\begin{aligned}\r\nP(-1) &= (-1)^6 - 1 \\\\\r\n&= 1 - 1 \\\\\r\n&= 0\r\n\\end{aligned}\r\n$$\r\nThe remainder is **0**. This also means $(x+1)$ is a factor of $x^6 - 1$.\r\n\r\n***\r\n\r\n#### Question 18: Finding a Value Using Remainder Theorem\r\n**The polynomial $f(x) = x^3 - (p - 2)x^2 - px + 28$ leaves a remainder of 10 when divided by $x + 3$. Find the value of $p$.**\r\n\r\n**Solution:**\r\nGiven $f(x) = x^3 - (p - 2)x^2 - px + 28$. The divisor is $(x+3)$, so $a=-3$. The remainder is $R=10$.\r\n\r\nBy the Remainder Theorem, $f(-3) = 10$.\r\n$$\r\n\\begin{aligned}\r\n(-3)^3 - (p - 2)(-3)^2 - p(-3) + 28 &= 10 \\\\\r\n-27 - (p - 2)(9) + 3p + 28 &= 10 \\\\\r\n-27 - (9p - 18) + 3p + 28 &= 10 \\\\\r\n-27 - 9p + 18 + 3p + 28 &= 10 \\\\\r\n19 - 6p &= 10 \\\\\r\n9 &= 6p \\\\\r\np &= \\frac{9}{6} = \\frac{3}{2}\r\n\\end{aligned}\r\n$$\r\nThus, **$p = \\frac{3}{2}$**.\r\n\r\n---\r\n\r\n### Long Questions (Solving Cubic Equations)\r\n\r\n***\r\n\r\n#### Question 19: Solve the Polynomial Equation\r\n**Solve: $2x^3 - 9x^2 + 7x + 6 = 0$**\r\n\r\n**Solution:**\r\nLet $P(x) = 2x^3 - 9x^2 + 7x + 6$. We test integer factors of the constant term (6).\r\nTest $x=2$:\r\n$$P(2) = 2(2)^3 - 9(2)^2 + 7(2) + 6 = 16 - 36 + 14 + 6 = 0$$\r\nSo, $(x-2)$ is a factor. We use synthetic division with $a=2$ to find the other factors.\r\n2 | 2 -9 7 6 | 4 -10 -6\r\n 2   -5   -3    0\r\nThe remaining quadratic is $2x^2 - 5x - 3$. We factor this:\r\n$$\r\n\\begin{aligned}\r\n2x^2 - 5x - 3 &= 0 \\\\\r\n2x^2 - 6x + x - 3 &= 0 \\\\\r\n2x(x - 3) + 1(x - 3) &= 0 \\\\\r\n(2x + 1)(x - 3) &= 0\r\n\\end{aligned}\r\n$$\r\nSo, the factors are $(x-2)$, $(x-3)$, and $(2x+1)$. The solutions are the roots of these factors.\r\nThus, **$x = 2$, $x = 3$, and $x = -\\frac{1}{2}$**.\r\n\r\n***\r\n\r\n#### Question 20: Solve the Polynomial Equation\r\n**Solve: $3x^3 - 7x^2 + 4 = 0$**\r\n\r\n**Solution:**\r\nLet $P(x) = 3x^3 - 7x^2 + 0x + 4$. Test factors of 4.\r\nTest $x=1$: $P(1) = 3(1)^3 - 7(1)^2 + 4 = 3 - 7 + 4 = 0$.\r\nSo, $(x-1)$ is a factor. Use synthetic division with $a=1$.\r\n1 | 3 -7 0 4 | 3 -4 -4\r\n 3   -4   -4    0\r\nThe remaining quadratic is $3x^2 - 4x - 4$. We factor this:\r\n$$\r\n\\begin{aligned}\r\n3x^2 - 4x - 4 &= 0 \\\\\r\n3x^2 - 6x + 2x - 4 &= 0 \\\\\r\n3x(x - 2) + 2(x - 2) &= 0 \\\\\r\n(3x + 2)(x - 2) &= 0\r\n\\end{aligned}\r\n$$\r\nThe roots are from $(x-1)$, $(x-2)$, and $(3x+2)$.\r\nThus, **$x = 1$, $x = 2$, and $x = -\\frac{2}{3}$**.\r\n\r\n***\r\n\r\n#### Question 21: Solve the Polynomial Equation\r\n**Solve: $y^3 - 6y^2 + 11y - 6 = 0$**\r\n\r\n**Solution:**\r\nLet $P(y) = y^3 - 6y^2 + 11y - 6$. Test factors of 6.\r\nTest $y=1$: $P(1) = 1 - 6 + 11 - 6 = 0$.\r\nSo, $(y-1)$ is a factor. Use synthetic division with $a=1$.\r\n1 | 1 -6 11 -6 | 1 -5 6\r\n 1   -5    6    0\r\nThe remaining quadratic is $y^2 - 5y + 6$. We factor this:\r\n$$\r\n\\begin{aligned}\r\ny^2 - 5y + 6 &= 0 \\\\\r\n(y-2)(y-3) &= 0\r\n\\end{aligned}\r\n$$\r\nThe roots are from $(y-1)$, $(y-2)$, and $(y-3)$.\r\nThus, **$y = 1, 2, 3$**.\r\n\r\n***\r\n\r\n#### Question 22: Solve the Polynomial Equation\r\n**Solve: $6x^3 + x^2 - 19x + 6 = 0$**\r\n\r\n**Solution:**\r\nLet $P(x) = 6x^3 + x^2 - 19x + 6$. Test factors of 6.\r\nTest $x=-2$: $P(-2) = 6(-8) + (-2)^2 - 19(-2) + 6 = -48 + 4 + 38 + 6 = 0$.\r\nSo, $(x+2)$ is a factor. Use synthetic division with $a=-2$.\r\n-2 | 6 1 -19 6 | -12 22 -6\r\n 6  -11     3    0\r\nThe remaining quadratic is $6x^2 - 11x + 3$. We factor this:\r\n$$\r\n\\begin{aligned}\r\n6x^2 - 11x + 3 &= 0 \\\\\r\n6x^2 - 9x - 2x + 3 &= 0 \\\\\r\n3x(2x - 3) - 1(2x - 3) &= 0 \\\\\r\n(3x - 1)(2x - 3) &= 0\r\n\\end{aligned}\r\n$$\r\nThe roots are from $(x+2)$, $(3x-1)$, and $(2x-3)$.\r\nThus, **$x = -2$, $x = \\frac{1}{3}$, and $x = \\frac{3}{2}$**.\r\n\r\n\r\n\r\n\r\n\r\n\r\nit must consist all concept of all these questions at first and then questions proper note for students to easy grasp concept and genrate this raw markdown code make sure it is not rendered here i want in code editor\r\n\r\n\r\n\r\n\r\nShow thinking\r\nOf course. Here is the complete study guide for the \"Polynomials\" chapter, designed with clear conceptual explanations first, followed by all the solutions with proper spacing. The entire note is provided as a single block of raw markdown code for you to copy and paste.\r\n\r\nMarkdown\r\n\r\n# Grade 10 Optional Mathematics: Polynomials Study Guide\r\n## Unit: Algebra | Chapter: Polynomials | Complete Solutions\r\n\r\n---\r\n\r\n## 🎯 Chapter Introduction\r\n\r\nWelcome to the comprehensive study guide for the **Polynomials** chapter. This note is designed to make complex concepts easy to understand, with clear definitions and detailed, step-by-step solutions to 35 different problems. By the end of this guide, you will have mastered the key theorems and techniques needed for your exams. 🧠\r\n\r\n### Key Learning Areas\r\n1.  **Remainder Theorem:** A quick way to find the remainder of a polynomial division.\r\n2.  **Factor Theorem:** A method to check if a linear expression is a factor of a polynomial.\r\n3.  **Synthetic Division:** A shortcut for dividing a polynomial by a linear factor.\r\n4.  **Solving Cubic Equations:** A systematic process to find the roots of cubic polynomials.\r\n\r\n---\r\n\r\n## 🔑 Key Concepts & Theorems\r\n\r\nBefore we solve the problems, let\'s understand the main tools we\'ll be using.\r\n\r\n### 1. The Remainder Theorem\r\n\r\n**What it is:** If you divide a polynomial, $P(x)$, by a linear expression of the form $(x - a)$, the remainder will be equal to the value of the polynomial when you substitute $a$ for $x$.\r\n\r\n* **In simple terms:** Remainder = $P(a)$.\r\n\r\n**How to use it:**\r\n1.  Identify your polynomial $P(x)$ and your divisor $(x - a)$.\r\n2.  Find the value of $a$. Remember, if the divisor is $(x+3)$, then $a = -3$.\r\n3.  Calculate $P(a)$ by substituting $a$ into the polynomial. The result is your remainder.\r\n\r\n---\r\n\r\n### 2. The Factor Theorem\r\n\r\n**What it is:** This is a special case of the Remainder Theorem. It gives you a way to check if $(x-a)$ is a factor of a polynomial $P(x)$.\r\n\r\n* **Rule:** $(x - a)$ is a factor of $P(x)$ **if and only if** $P(a) = 0$.\r\n\r\n**How to use it:**\r\n1.  To check if $(x-a)$ is a factor, calculate the remainder $P(a)$.\r\n2.  If the remainder is 0, it **is** a factor.\r\n3.  If the remainder is not 0, it **is not** a factor.\r\n\r\n---\r\n\r\n### 3. Synthetic Division\r\n\r\n**What it is:** A fast, shorthand method for dividing a polynomial by a linear factor of the form $(x-a)$. It is used to find the quotient and the remainder. \r\n\r\n**How to use it:**\r\n1.  Write down the coefficients of the polynomial. **Important:** If any term is missing (like an $x^2$ term in a cubic), you must use a **0** as a placeholder for its coefficient.\r\n2.  Write the value of $a$ from the divisor $(x-a)$ to the left.\r\n3.  Bring down the first coefficient.\r\n4.  Multiply this number by $a$ and write the result under the next coefficient.\r\n5.  Add the numbers in that column.\r\n6.  Repeat steps 4 and 5 until you reach the end.\r\n\r\nThe last number is the remainder, and the other numbers are the coefficients of the quotient.\r\n\r\n**Example:** Divide $x^3 - 7x - 6$ by $(x+2)$. Here, $a=-2$. The coefficients are 1, 0, -7, -6.\r\n-2 | 1 0 -7 -6 | -2 4 6\r\n 1  -2   -3    0  <-- Remainder is 0\r\nThe quotient is $x^2 - 2x - 3$.\r\n\r\n---\r\n\r\n### 4. Solving Cubic Equations\r\n\r\n**What it is:** Finding the values of $x$ that make a cubic polynomial equal to zero. A cubic equation has three roots.\r\n\r\n**How to solve ($ax^3 + bx^2 + cx + d = 0$):**\r\n1.  **Find one factor:** Use the Factor Theorem. Test integer factors of the constant term, $d$. Find a value \'a\' for which $P(a) = 0$. This gives you one factor, $(x-a)$.\r\n2.  **Reduce the polynomial:** Use synthetic division to divide the polynomial by the factor you just found. The result will be a quadratic expression.\r\n3.  **Solve the quadratic:** Solve the resulting quadratic equation using factoring or the quadratic formula to find the other two roots.\r\n4.  **List all roots:** Combine the root from Step 1 with the two roots from Step 3.\r\n\r\n---\r\n\r\n## 📚 Exercise Solutions\r\n\r\n### Short Questions\r\n\r\n***\r\n\r\n#### Question 1: Finding a Value Using Factor Theorem\r\n**If $x - 2$ is a factor of $2x^3 - 6x^2 - 5m - 2$, find the value of $m$.**\r\n\r\n**Solution:**\r\nLet the polynomial be $P(x) = 2x^3 - 6x^2 - 5m - 2$.\r\nThe divisor is $d(x) = x - 2$. Comparing this to $x - a$, we get $a = 2$.\r\n\r\nBy the Factor Theorem, if $(x - a)$ is a factor of $P(x)$, then $P(a) = 0$.\r\n$$\r\n\\begin{aligned}\r\nP(2) &= 0 \\\\\r\n2(2)^3 - 6(2)^2 - 5m - 2 &= 0 \\\\\r\n2(8) - 6(4) - 5m - 2 &= 0 \\\\\r\n16 - 24 - 5m - 2 &= 0 \\\\\r\n-10 - 5m &= 0 \\\\\r\n-10 &= 5m \\\\\r\nm &= -2\r\n\\end{aligned}\r\n$$\r\nThus, **$m = -2$**.\r\n\r\n***\r\n\r\n#### Question 2: Finding a Value Using Factor Theorem\r\n**If $(x + 2)$ is a factor of the polynomial $2x^3 + 3x^2 - mx + 4$, find the value of $m$.**\r\n\r\n**Solution:**\r\nLet $P(x) = 2x^3 + 3x^2 - mx + 4$.\r\nThe divisor is $d(x) = x + 2$. Comparing this to $x - a$, we get $a = -2$.\r\n\r\nBy the Factor Theorem, $P(-2) = 0$.\r\n$$\r\n\\begin{aligned}\r\n2(-2)^3 + 3(-2)^2 - m(-2) + 4 &= 0 \\\\\r\n2(-8) + 3(4) + 2m + 4 &= 0 \\\\\r\n-16 + 12 + 2m + 4 &= 0 \\\\\r\n0 + 2m &= 0 \\\\\r\n2m &= 0 \\\\\r\nm &= 0\r\n\\end{aligned}\r\n$$\r\nThus, **$m = 0$**.\r\n\r\n***\r\n\r\n#### Question 3: Finding a Value Using Remainder Theorem\r\n**If $2x^3 - 4x^2 + kx + 10$ is divided by $(x + 2)$, the remainder is 4. Find the value of $k$.**\r\n\r\n**Solution:**\r\nLet $P(x) = 2x^3 - 4x^2 + kx + 10$.\r\nThe divisor is $d(x) = x + 2$, so $a = -2$. The remainder is $R = 4$.\r\n\r\nBy the Remainder Theorem, $P(a) = R$.\r\n$$\r\n\\begin{aligned}\r\nP(-2) &= 4 \\\\\r\n2(-2)^3 - 4(-2)^2 + k(-2) + 10 &= 4 \\\\\r\n2(-8) - 4(4) - 2k + 10 &= 4 \\\\\r\n-16 - 16 - 2k + 10 &= 4 \\\\\r\n-22 - 2k &= 4 \\\\\r\n-2k &= 26 \\\\\r\nk &= -13\r\n\\end{aligned}\r\n$$\r\nThus, **$k = -13$**.\r\n\r\n***\r\n\r\n#### Question 4: Finding a Value Using Factor Theorem\r\n**If $x - 3$ is a factor of $2x^3 - x^2 + 10x - k$, find the value of $k$.**\r\n\r\n**Solution:**\r\nLet $P(x) = 2x^3 - x^2 + 10x - k$.\r\nThe divisor is $d(x) = x - 3$, so $a = 3$.\r\n\r\nBy the Factor Theorem, $P(3) = 0$.\r\n$$\r\n\\begin{aligned}\r\n2(3)^3 - (3)^2 + 10(3) - k &= 0 \\\\\r\n2(27) - 9 + 30 - k &= 0 \\\\\r\n54 - 9 + 30 - k &= 0 \\\\\r\n75 - k &= 0 \\\\\r\nk &= 75\r\n\\end{aligned}\r\n$$\r\nThus, **$k = 75$**.\r\n\r\n***\r\n\r\n#### Question 5: Finding a Value Using Factor Theorem\r\n**If $x - 5$ is a factor of $x^3 + px^2 + 4x + 5$, find the value of $p$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 + px^2 + 4x + 5$.\r\nThe divisor is $d(x) = x - 5$, so $a = 5$.\r\n\r\nBy the Factor Theorem, $P(5) = 0$.\r\n$$\r\n\\begin{aligned}\r\n(5)^3 + p(5)^2 + 4(5) + 5 &= 0 \\\\\r\n125 + 25p + 20 + 5 &= 0 \\\\\r\n150 + 25p &= 0 \\\\\r\n25p &= -150 \\\\\r\np &= -6\r\n\\end{aligned}\r\n$$\r\nThus, **$p = -6$**.\r\n\r\n***\r\n\r\n#### Question 6: Finding the Remainder\r\n**Using the Remainder Theorem, find the remainder when $8x^3 - 4x^2 + 2x - 5$ is divided by $2x - 1$.**\r\n\r\n**Solution:**\r\nLet $P(x) = 8x^3 - 4x^2 + 2x - 5$.\r\nFor the divisor $2x - 1$, we set it to zero to find the value to substitute: $2x - 1 = 0 \\implies x = \\frac{1}{2}$.\r\n\r\nBy the Remainder Theorem, the remainder is $P(\\frac{1}{2})$.\r\n$$\r\n\\begin{aligned}\r\nP\\left(\\frac{1}{2}\\right) &= 8\\left(\\frac{1}{2}\\right)^3 - 4\\left(\\frac{1}{2}\\right)^2 + 2\\left(\\frac{1}{2}\\right) - 5 \\\\\r\n&= 8\\left(\\frac{1}{8}\\right) - 4\\left(\\frac{1}{4}\\right) + 1 - 5 \\\\\r\n&= 1 - 1 + 1 - 5 \\\\\r\n&= -4\r\n\\end{aligned}\r\n$$\r\nThe remainder is **-4**.\r\n\r\n***\r\n\r\n#### Question 7: Using the Factor Theorem\r\n**If $x - 5$ is a factor of $2x^3 - 7x + (p - 12)$, find the value of p.**\r\n\r\n**Solution:**\r\nLet $P(x) = 2x^3 - 7x + p - 12$.\r\nSince $x-5$ is a factor, $a=5$ and $P(5)=0$.\r\n$$\r\n\\begin{aligned}\r\nP(5) &= 0 \\\\\r\n2(5)^3 - 7(5) + p - 12 &= 0 \\\\\r\n2(125) - 35 + p - 12 &= 0 \\\\\r\n250 - 35 + p - 12 &= 0 \\\\\r\n203 + p &= 0 \\\\\r\np &= -203\r\n\\end{aligned}\r\n$$\r\nThus, **$p = -203$**.\r\n\r\n***\r\n\r\n#### Question 8: Using Synthetic Division\r\n**If $x^3 - 21x - 20 = (x + 1) \\cdot q(x)$, find $q(x)$ by using synthetic division.**\r\n\r\n**Solution:**\r\nThe polynomial is $P(x) = x^3 + 0x^2 - 21x - 20$.\r\nThe divisor is $x+1$, so we use $a = -1$ for synthetic division.\r\n\r\n-1 | 1 0 -21 -20 | -1 1 20\r\n 1  -1   -20     0 \r\n\r\nThe quotient\'s coefficients are 1, -1, and -20. The remainder is 0.\r\nThus, **$q(x) = x^2 - x - 20$**.\r\n\r\n***\r\n\r\n#### Question 9: Using Synthetic Division\r\n**If $x^3 - 19x - 30 = (x + 2) \\cdot q(x)$, find $q(x)$ by using synthetic division.**\r\n\r\n**Solution:**\r\nThe polynomial is $P(x) = x^3 + 0x^2 - 19x - 30$.\r\nThe divisor is $x+2$, so we use $a = -2$.\r\n\r\n-2 | 1 0 -19 -30 | -2 4 30\r\n 1  -2   -15     0 \r\n\r\nThe quotient\'s coefficients are 1, -2, and -15. The remainder is 0.\r\nThus, **$q(x) = x^2 - 2x - 15$**.\r\n\r\n***\r\n\r\n#### Question 10: State and Apply the Factor Theorem\r\n**State the Factor Theorem. Use it to determine whether $x + 3$ is a factor of $x^3 - 8x + 3$.**\r\n\r\n**Solution:**\r\n**Factor Theorem:** A polynomial $P(x)$ has a factor $(x - a)$ if and only if $P(a) = 0$.\r\n\r\nLet $P(x) = x^3 - 8x + 3$. To check if $(x+3)$ is a factor, we test $P(-3)$.\r\n$$\r\n\\begin{aligned}\r\nP(-3) &= (-3)^3 - 8(-3) + 3 \\\\\r\n&= -27 + 24 + 3 \\\\\r\n&= 0\r\n\\end{aligned}\r\n$$\r\nSince $P(-3) = 0$, **yes, $(x+3)$ is a factor** of the polynomial.\r\n\r\n***\r\n\r\n#### Question 11: State and Apply the Remainder Theorem\r\n**State the Remainder Theorem. When $x^3 + 5x^2 - 6x - 16$ is divided by $(x + 2)$, find the remainder.**\r\n\r\n**Solution:**\r\n**Remainder Theorem:** If a polynomial $P(x)$ is divided by $(x - a)$, the remainder is $P(a)$.\r\n\r\nLet $P(x) = x^3 + 5x^2 - 6x - 16$. The divisor is $(x+2)$, so $a = -2$. The remainder is $P(-2)$.\r\n$$\r\n\\begin{aligned}\r\nP(-2) &= (-2)^3 + 5(-2)^2 - 6(-2) - 16 \\\\\r\n&= -8 + 5(4) + 12 - 16 \\\\\r\n&= -8 + 20 + 12 - 16 \\\\\r\n&= 8\r\n\\end{aligned}\r\n$$\r\nThe remainder is **8**.\r\n\r\n***\r\n\r\n#### Question 12: Finding a Value Using Remainder Theorem\r\n**If the polynomial $x^3 + 6x^2 + kx + 10$ is divided by $(x + 2)$, the remainder is 4. Find the value of $k$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 + 6x^2 + kx + 10$. The divisor is $(x+2)$, so $a=-2$. The remainder is $R=4$.\r\n\r\nBy the Remainder Theorem, $P(-2) = 4$.\r\n$$\r\n\\begin{aligned}\r\n(-2)^3 + 6(-2)^2 + k(-2) + 10 &= 4 \\\\\r\n-8 + 6(4) - 2k + 10 &= 4 \\\\\r\n-8 + 24 - 2k + 10 &= 4 \\\\\r\n26 - 2k &= 4 \\\\\r\n22 &= 2k \\\\\r\nk &= 11\r\n\\end{aligned}\r\n$$\r\nThus, **$k = 11$**.\r\n\r\n***\r\n\r\n#### Question 13: Finding a Value Using Factor Theorem\r\n**If $(x + 2)$ is a factor of $x^3 - 19x - p$, find the value of $p$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 - 19x - p$. Since $(x+2)$ is a factor, $a=-2$ and $P(-2)=0$.\r\n$$\r\n\\begin{aligned}\r\nP(-2) &= 0 \\\\\r\n(-2)^3 - 19(-2) - p &= 0 \\\\\r\n-8 + 38 - p &= 0 \\\\\r\n30 - p &= 0 \\\\\r\np &= 30\r\n\\end{aligned}\r\n$$\r\nThus, **$p = 30$**.\r\n\r\n***\r\n\r\n#### Question 14: Finding a Value Using Factor Theorem\r\n**If $(x - m)$ is a factor of $x^3 - mx^2 - 4x + m + 12$, calculate the value of $m$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 - mx^2 - 4x + m + 12$. Since $(x-m)$ is a factor, $a=m$ and $P(m)=0$.\r\n$$\r\n\\begin{aligned}\r\nP(m) &= 0 \\\\\r\n(m)^3 - m(m)^2 - 4(m) + m + 12 &= 0 \\\\\r\nm^3 - m^3 - 4m + m + 12 &= 0 \\\\\r\n-3m + 12 &= 0 \\\\\r\n12 &= 3m \\\\\r\nm &= 4\r\n\\end{aligned}\r\n$$\r\nThus, **$m = 4$**.\r\n\r\n***\r\n\r\n#### Question 15: Finding a Value Using Factor Theorem\r\n**If $x - 2$ is a factor of $x^3 - kx^2 + x - 6$, find the value of $k$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 - kx^2 + x - 6$. Since $(x-2)$ is a factor, $a=2$ and $P(2)=0$.\r\n$$\r\n\\begin{aligned}\r\nP(2) &= 0 \\\\\r\n(2)^3 - k(2)^2 + 2 - 6 &= 0 \\\\\r\n8 - 4k - 4 &= 0 \\\\\r\n4 - 4k &= 0 \\\\\r\n4 &= 4k \\\\\r\nk &= 1\r\n\\end{aligned}\r\n$$\r\nThus, **$k=1$**.\r\n\r\n***\r\n\r\n#### Question 16: Using the Factor Theorem\r\n**Use the Factor Theorem to determine whether $x + 4$ is a factor of the polynomial $x^3 - 15x + 4$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^3 - 15x + 4$. To check if $(x+4)$ is a factor, we test if $P(-4) = 0$.\r\n$$\r\n\\begin{aligned}\r\nP(-4) &= (-4)^3 - 15(-4) + 4 \\\\\r\n&= -64 + 60 + 4 \\\\\r\n&= 0\r\n\\end{aligned}\r\n$$\r\nSince $P(-4) = 0$, **yes, $(x+4)$ is a factor**.\r\n\r\n***\r\n\r\n#### Question 17: Using the Remainder Theorem\r\n**Find the remainder when the polynomial $x^6 - 1$ is divided by $x + 1$.**\r\n\r\n**Solution:**\r\nLet $P(x) = x^6 - 1$. The divisor is $(x+1)$, so $a=-1$. The remainder is $P(-1)$.\r\n$$\r\n\\begin{aligned}\r\nP(-1) &= (-1)^6 - 1 \\\\\r\n&= 1 - 1 \\\\\r\n&= 0\r\n\\end{aligned}\r\n$$\r\nThe remainder is **0**. This also means $(x+1)$ is a factor of $x^6 - 1$.\r\n\r\n***\r\n\r\n#### Question 18: Finding a Value Using Remainder Theorem\r\n**The polynomial $f(x) = x^3 - (p - 2)x^2 - px + 28$ leaves a remainder of 10 when divided by $x + 3$. Find the value of $p$.**\r\n\r\n**Solution:**\r\nGiven $f(x) = x^3 - (p - 2)x^2 - px + 28$. The divisor is $(x+3)$, so $a=-3$. The remainder is $R=10$.\r\n\r\nBy the Remainder Theorem, $f(-3) = 10$.\r\n$$\r\n\\begin{aligned}\r\n(-3)^3 - (p - 2)(-3)^2 - p(-3) + 28 &= 10 \\\\\r\n-27 - (p - 2)(9) + 3p + 28 &= 10 \\\\\r\n-27 - (9p - 18) + 3p + 28 &= 10 \\\\\r\n-27 - 9p + 18 + 3p + 28 &= 10 \\\\\r\n19 - 6p &= 10 \\\\\r\n9 &= 6p \\\\\r\np &= \\frac{9}{6} = \\frac{3}{2}\r\n\\end{aligned}\r\n$$\r\nThus, **$p = \\frac{3}{2}$**.\r\n\r\n---\r\n\r\n### Long Questions (Solving Cubic Equations)\r\n\r\n***\r\n\r\n#### Question 19: Solve the Polynomial Equation\r\n**Solve: $2x^3 - 9x^2 + 7x + 6 = 0$**\r\n\r\n**Solution:**\r\nLet $P(x) = 2x^3 - 9x^2 + 7x + 6$. Test integer factors of 6.\r\nWhen $x=2$, $P(2) = 2(8) - 9(4) + 7(2) + 6 = 16 - 36 + 14 + 6 = 0$.\r\nSo, $(x-2)$ is a factor. Use synthetic division with $a=2$.\r\n2 | 2 -9 7 6 | 4 -10 -6\r\n 2   -5   -3    0\r\nThe remaining quadratic is $2x^2 - 5x - 3$. Factor it:\r\n$$\r\n\\begin{aligned}\r\n2x^2 - 5x - 3 &= 0 \\\\\r\n2x^2 - 6x + x - 3 &= 0 \\\\\r\n2x(x - 3) + 1(x - 3) &= 0 \\\\\r\n(2x + 1)(x - 3) &= 0\r\n\\end{aligned}\r\n$$\r\nThe factors are $(x-2)$, $(x-3)$, and $(2x+1)$.\r\nThus, the solutions are **$x = 2$, $x = 3$, and $x = -\\frac{1}{2}$**.\r\n\r\n***\r\n\r\n#### Question 20: Solve the Polynomial Equation\r\n**Solve: $3x^3 - 7x^2 + 4 = 0$**\r\n\r\n**Solution:**\r\nLet $P(x) = 3x^3 - 7x^2 + 0x + 4$. Test factors of 4.\r\nWhen $x=1$, $P(1) = 3 - 7 + 4 = 0$.\r\nSo, $(x-1)$ is a factor. Use synthetic division with $a=1$.\r\n1 | 3 -7 0 4 | 3 -4 -4\r\n 3   -4   -4    0\r\nThe remaining quadratic is $3x^2 - 4x - 4$. Factor it:\r\n$$\r\n\\begin{aligned}\r\n3x^2 - 4x - 4 &= 0 \\\\\r\n3x^2 - 6x + 2x - 4 &= 0 \\\\\r\n3x(x - 2) + 2(x - 2) &= 0 \\\\\r\n(3x + 2)(x - 2) &= 0\r\n\\end{aligned}\r\n$$\r\nThe solutions are from $(x-1)$, $(x-2)$, and $(3x+2)$.\r\nThus, **$x = 1$, $x = 2$, and $x = -\\frac{2}{3}$**.\r\n\r\n***\r\n\r\n#### Question 21: Solve the Polynomial Equation\r\n**Solve: $y^3 - 6y^2 + 11y - 6 = 0$**\r\n\r\n**Solution:**\r\nLet $P(y) = y^3 - 6y^2 + 11y - 6$. Test factors of 6.\r\nWhen $y=1$, $P(1) = 1 - 6 + 11 - 6 = 0$.\r\nSo, $(y-1)$ is a factor. Use synthetic division with $a=1$.\r\n1 | 1 -6 11 -6 | 1 -5 6\r\n 1   -5    6    0\r\nThe remaining quadratic is $y^2 - 5y + 6$. Factor it:\r\n$$\r\ny^2 - 5y + 6 = (y-2)(y-3) = 0\r\n$$\r\nThe roots are from $(y-1)$, $(y-2)$, and $(y-3)$.\r\nThus, **$y = 1, 2, 3$**.\r\n\r\n***\r\n\r\n#### Question 22: Solve the Polynomial Equation\r\n**Solve: $6x^3 + x^2 - 19x + 6 = 0$**\r\n\r\n**Solution:**\r\nLet $P(x) = 6x^3 + x^2 - 19x + 6$. Test factors of 6.\r\nWhen $x=-2$, $P(-2) = 6(-8) + 4 - 19(-2) + 6 = -48 + 4 + 38 + 6 = 0$.\r\nSo, $(x+2)$ is a factor. Use synthetic division with $a=-2$.\r\n-2 | 6 1 -19 6 | -12 22 -6\r\n 6  -11     3    0\r\nThe remaining quadratic is $6x^2 - 11x + 3$. Factor it:\r\n$$\r\n\\begin{aligned}\r\n6x^2 - 11x + 3 &= 0 \\\\\r\n6x^2 - 9x - 2x + 3 &= 0 \\\\\r\n3x(2x - 3) - 1(2x - 3) &= 0 \\\\\r\n(3x - 1)(2x - 3) &= 0\r\n\\end{aligned}\r\n$$\r\nThe roots are from $(x+2)$, $(3x-1)$, and $(2x-3)$.\r\nThus, **$x = -2$, $x = \\frac{1}{3}$, and $x = \\frac{3}{2}$**.', 1, '2025-07-25 01:11:38', '2025-09-18 07:21:49');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(140, 13103, '# Grade 10 Optional Mathematics: Sequence and Series Study Guide\r\n## Unit: Algebra | Chapter: Sequence and Series | Complete Solutions\r\n\r\n---\r\n\r\n## 🎯 Chapter Introduction\r\n\r\nWelcome to the comprehensive study guide for the **Sequence and Series** chapter. This note is designed to make complex concepts easy to understand, with clear definitions, essential formulas, and detailed, step-by-step solutions to a wide range of problems. By the end of this guide, you will have mastered Arithmetic and Geometric progressions. 📈\r\n\r\n### Key Learning Areas\r\n1.  **Definitions:** Understanding Sequences, Series, and their different types.\r\n2.  **Arithmetic Progression (A.P.):** Working with sequences that have a common difference.\r\n3.  **Geometric Progression (G.P.):** Working with sequences that have a common ratio.\r\n4.  **Means:** Calculating and inserting Arithmetic and Geometric Means.\r\n5.  **Summation:** Finding the sum of a given number of terms in a series.\r\n\r\n---\r\n\r\n## 🔑 Key Concepts & Formulas\r\n\r\nBefore solving the problems, let\'s review the fundamental concepts and formulas.\r\n\r\n### 1. Core Definitions\r\n\r\n* **Sequence:** A set of numbers arranged in a definite order according to a specific mathematical rule.\r\n    * *Example:* `1, 3, 5, 7, ...` is a sequence where each term is 2 more than the previous one.\r\n\r\n* **Series:** The sum of the terms in a sequence.\r\n    * *Example:* `1 + 3 + 5 + 7 + ...` is the series corresponding to the sequence above.\r\n\r\n* **Arithmetic Progression (A.P.):** A sequence where the difference between any two consecutive terms is constant. This constant is called the **common difference (d)**.\r\n    * *Example:* `5, 9, 13, 17, ...` (common difference is 4).\r\n\r\n* **Geometric Progression (G.P.):** A sequence where the ratio of any term to its preceding term is constant. This constant is called the **common ratio (r)**.\r\n    * *Example:* `3, 6, 12, 24, ...` (common ratio is 2).\r\n\r\n* **Arithmetic Mean (A.M.):** A term (or terms) inserted between two given terms that maintains the arithmetic progression.\r\n\r\n* **Geometric Mean (G.M.):** A term (or terms) inserted between two given terms that maintains the geometric progression.\r\n\r\n---\r\n\r\n### 2. Formulas for Arithmetic Progression (A.P.)\r\n\r\nLet **a** be the first term, **d** be the common difference, and **n** be the number of terms.\r\n\r\n* **nth Term ($t_n$):**\r\n    $$t_n = a + (n-1)d$$\r\n\r\n* **Sum of First n Terms ($S_n$):**\r\n    $$S_n = \\frac{n}{2}[2a + (n-1)d]$$\r\n\r\n* **Arithmetic Mean between a and b:**\r\n    $$A.M. = \\frac{a+b}{2}$$\r\n\r\n* **Inserting \'n\' Arithmetic Means between a and b:**\r\n    The common difference `d` is found using:\r\n    $$d = \\frac{b-a}{n+1}$$\r\n\r\n---\r\n\r\n### 3. Formulas for Geometric Progression (G.P.)\r\n\r\nLet **a** be the first term, **r** be the common ratio, and **n** be the number of terms.\r\n\r\n* **nth Term ($t_n$):**\r\n    $$t_n = ar^{n-1}$$\r\n\r\n* **Sum of First n Terms ($S_n$):**\r\n    $$S_n = \\frac{a(r^n - 1)}{r-1} \\quad (\\text{if } r > 1)$$\r\n    $$S_n = \\frac{a(1 - r^n)}{1-r} \\quad (\\text{if } r < 1)$$\r\n\r\n* **Geometric Mean between a and b:**\r\n    $$G.M. = \\sqrt{ab}$$\r\n\r\n* **Inserting \'n\' Geometric Means between a and b:**\r\n    The common ratio `r` is found using:\r\n    $$r = \\left(\\frac{b}{a}\\right)^{\\frac{1}{n+1}}$$\r\n\r\n---\r\n\r\n## 📚 Exercise Solutions\r\n\r\n### Short Questions\r\n\r\n***\r\n\r\n#### Question 1: Arithmetic and Geometric Mean\r\n**Find the arithmetic mean and geometric mean between 4 and 16.**\r\n\r\n**Solution:**\r\nHere, $a = 4$ and $b = 16$.\r\n\r\n**Arithmetic Mean (A.M.):**\r\n$$\r\n\\begin{aligned}\r\nA.M. &= \\frac{a+b}{2} \\\\\r\n&= \\frac{4+16}{2} \\\\\r\n&= \\frac{20}{2} = 10\r\n\\end{aligned}\r\n$$\r\n**Geometric Mean (G.M.):**\r\n$$\r\n\\begin{aligned}\r\nG.M. &= \\sqrt{ab} \\\\\r\n&= \\sqrt{4 \\times 16} \\\\\r\n&= \\sqrt{64} = 8\r\n\\end{aligned}\r\n$$\r\nThe A.M. is **10** and the G.M. is **8**.\r\n\r\n***\r\n\r\n#### Question 2: Arithmetic and Geometric Mean\r\n**Find the arithmetic mean and geometric mean between 3 and 27.**\r\n\r\n**Solution:**\r\nHere, $a = 3$ and $b = 27$.\r\n\r\n**Arithmetic Mean (A.M.):**\r\n$$\r\nA.M. = \\frac{3+27}{2} = \\frac{30}{2} = 15\r\n$$\r\n**Geometric Mean (G.M.):**\r\n$$\r\nG.M. = \\sqrt{3 \\times 27} = \\sqrt{81} = 9\r\n$$\r\nThe A.M. is **15** and the G.M. is **9**.\r\n\r\n***\r\n\r\n#### Question 3: Geometric Sequence\r\n**If 4, a, and 16 are in a geometric sequence, find the value of a.**\r\n\r\n**Solution:**\r\nIn a geometric sequence, the middle term is the geometric mean of the first and last terms.\r\n$$\r\n\\begin{aligned}\r\na &= \\sqrt{4 \\times 16} \\\\\r\na &= \\sqrt{64} \\\\\r\na &= 8\r\n\\end{aligned}\r\n$$\r\nAlternatively, the ratio of consecutive terms is constant:\r\n$$\r\n\\begin{aligned}\r\n\\frac{a}{4} &= \\frac{16}{a} \\\\\r\na^2 &= 64 \\\\\r\na &= 8\r\n\\end{aligned}\r\n$$\r\nThus, **$a=8$**.\r\n\r\n***\r\n\r\n#### Question 4: Relation between Means\r\n**If the arithmetic mean between 4 and x is 34, find their geometric mean.**\r\n\r\n**Solution:**\r\nFirst, use the A.M. formula to find $x$.\r\n$$\r\n\\begin{aligned}\r\nA.M. &= \\frac{4+x}{2} \\\\\r\n34 &= \\frac{4+x}{2} \\\\\r\n68 &= 4+x \\\\\r\nx &= 64\r\n\\end{aligned}\r\n$$\r\nNow, find the G.M. between 4 and 64.\r\n$$\r\nG.M. = \\sqrt{4 \\times 64} = \\sqrt{256} = 16\r\n$$\r\nThe geometric mean is **16**.\r\n\r\n***\r\n\r\n#### Question 5: Relation between Means\r\n**If the geometric mean of 2 and x is 4, find their arithmetic mean.**\r\n\r\n**Solution:**\r\nFirst, use the G.M. formula to find $x$.\r\n$$\r\n\\begin{aligned}\r\nG.M. &= \\sqrt{2x} \\\\\r\n4 &= \\sqrt{2x} \\\\\r\n16 &= 2x \\\\\r\nx &= 8\r\n\\end{aligned}\r\n$$\r\nNow, find the A.M. between 2 and 8.\r\n$$\r\nA.M. = \\frac{2+8}{2} = \\frac{10}{2} = 5\r\n$$\r\nThe arithmetic mean is **5**.\r\n\r\n***\r\n\r\n#### Question 6: Geometric Sequence\r\n**If $m + 2$, $m + 8$, and $17 + m$ are in geometric sequence, find the value of m.**\r\n\r\n**Solution:**\r\nSince the terms are in a G.P., the ratio of consecutive terms is constant.\r\n$$\r\n\\begin{aligned}\r\n\\frac{m+8}{m+2} &= \\frac{17+m}{m+8} \\\\\r\n(m+8)^2 &= (m+2)(17+m) \\\\\r\nm^2 + 16m + 64 &= 17m + m^2 + 34 + 2m \\\\\r\nm^2 + 16m + 64 &= m^2 + 19m + 34 \\\\\r\n16m + 64 &= 19m + 34 \\\\\r\n30 &= 3m \\\\\r\nm &= 10\r\n\\end{aligned}\r\n$$\r\nThus, **$m=10$**.\r\n\r\n***\r\n\r\n#### Question 7: Finding the Term Number in an A.P.\r\n**Which term of the series $5 + 9 + 13 + ...$ is 85?**\r\n\r\n**Solution:**\r\nThis is an arithmetic series with:\r\nFirst term, $a = 5$.\r\nCommon difference, $d = 9 - 5 = 4$.\r\nLet the nth term, $t_n = 85$.\r\n\r\nUsing the formula $t_n = a + (n-1)d$:\r\n$$\r\n\\begin{aligned}\r\n85 &= 5 + (n-1)4 \\\\\r\n80 &= (n-1)4 \\\\\r\n20 &= n-1 \\\\\r\nn &= 21\r\n\\end{aligned}\r\n$$\r\nThus, 85 is the **21st term**.\r\n\r\n***\r\n\r\n#### Question 8: Finding the Term Number in a G.P.\r\n**Which term of the series $3 + 6 + 12 + ...$ is 192?**\r\n\r\n**Solution:**\r\nThis is a geometric series with:\r\nFirst term, $a = 3$.\r\nCommon ratio, $r = 6/3 = 2$.\r\nLet the nth term, $t_n = 192$.\r\n\r\nUsing the formula $t_n = ar^{n-1}$:\r\n$$\r\n\\begin{aligned}\r\n192 &= 3 \\cdot (2)^{n-1} \\\\\r\n64 &= (2)^{n-1} \\\\\r\n2^6 &= 2^{n-1} \\\\\r\n6 &= n-1 \\\\\r\nn &= 7\r\n\\end{aligned}\r\n$$\r\nThus, 192 is the **7th term**.\r\n\r\n***\r\n\r\n#### Question 9: Sum of a G.P.\r\n**Find the sum of the series $9 + 3 + 1 + ... + \\frac{1}{27}$.**\r\n\r\n**Solution:**\r\nThis is a geometric series with $a=9$ and $r = 3/9 = 1/3$. The last term is $t_n = 1/27$.\r\n\r\n**Step 1: Find the number of terms (n).**\r\n$$\r\n\\begin{aligned}\r\nt_n &= ar^{n-1} \\\\\r\n\\frac{1}{27} &= 9 \\cdot \\left(\\frac{1}{3}\\right)^{n-1} \\\\\r\n\\frac{1}{243} &= \\left(\\frac{1}{3}\\right)^{n-1} \\\\\r\n\\left(\\frac{1}{3}\\right)^5 &= \\left(\\frac{1}{3}\\right)^{n-1} \\\\\r\n5 &= n-1 \\\\\r\nn &= 6\r\n\\end{aligned}\r\n$$\r\n**Step 2: Find the sum of the 6 terms ($S_6$).**\r\nSince $r < 1$, we use $S_n = \\frac{a(1-r^n)}{1-r}$.\r\n$$\r\n\\begin{aligned}\r\nS_6 &= \\frac{9(1 - (1/3)^6)}{1 - 1/3} \\\\\r\n&= \\frac{9(1 - 1/729)}{2/3} \\\\\r\n&= \\frac{9(728/729)}{2/3} \\\\\r\n&= 9 \\times \\frac{728}{729} \\times \\frac{3}{2} \\\\\r\n&= \\frac{364}{27}\r\n\\end{aligned}\r\n$$\r\nThe sum of the series is **$\\frac{364}{27}$**.\r\n\r\n***\r\n\r\n#### Question 10: Product of Terms in a G.P.\r\n**If the third term of a geometric series is 3, find the product of the first five terms.**\r\n\r\n**Solution:**\r\nLet the first five terms of the G.P. be $a, ar, ar^2, ar^3, ar^4$.\r\nWe are given that the third term, $t_3 = ar^2 = 3$.\r\n\r\nThe product of the first five terms is:\r\n$$\r\n\\begin{aligned}\r\nP_5 &= (a)(ar)(ar^2)(ar^3)(ar^4) \\\\\r\n&= a^5 r^{0+1+2+3+4} \\\\\r\n&= a^5 r^{10} \\\\\r\n&= (ar^2)^5\r\n\\end{aligned}\r\n$$\r\nSince we know $ar^2 = 3$:\r\n$$\r\nP_5 = (3)^5 = 243\r\n$$\r\nThe product is **243**.\r\n\r\n***\r\n\r\n#### Question 11: Finding Terms in a G.P.\r\n**If $\\frac{1}{3}, p, q, 9$ are in a geometric sequence, find the values of p and q.**\r\n\r\n**Solution:**\r\nHere, $a = 1/3$ and the 4th term, $t_4 = 9$.\r\nFirst, find the common ratio $r$.\r\n$$\r\n\\begin{aligned}\r\nt_4 &= ar^{4-1} \\\\\r\n9 &= \\left(\\frac{1}{3}\\right)r^3 \\\\\r\n27 &= r^3 \\\\\r\nr &= 3\r\n\\end{aligned}\r\n$$\r\nNow find $p$ and $q$:\r\n$p$ is the 2nd term: $p = ar = (\\frac{1}{3})(3) = 1$.\r\n$q$ is the 3rd term: $q = ar^2 = (\\frac{1}{3})(3)^2 = 3$.\r\n\r\nThus, **$p=1$ and $q=3$**.\r\n\r\n***\r\n\r\n#### Question 12: Finding Common Ratio\r\n**Find the common ratio of a geometric sequence, whose first term is 2 and the third term is 242.**\r\n\r\n**Solution:**\r\nGiven $a=2$ and $t_3 = 242$.\r\n$$\r\n\\begin{aligned}\r\nt_3 &= ar^{3-1} \\\\\r\n242 &= 2 \\cdot r^2 \\\\\r\n121 &= r^2 \\\\\r\nr &= \\pm 11\r\n\\end{aligned}\r\n$$\r\nThe common ratio is **$\\pm 11$**.\r\n\r\n***\r\n\r\n#### Question 13: Finding Terms in an A.P.\r\n**If $5, x, y, 11$ is an arithmetic series. Find the value of x and y.**\r\n\r\n**Solution:**\r\nHere, $a = 5$ and the 4th term, $t_4 = 11$.\r\nFirst, find the common difference $d$.\r\n$$\r\n\\begin{aligned}\r\nt_4 &= a + (4-1)d \\\\\r\n11 &= 5 + 3d \\\\\r\n6 &= 3d \\\\\r\nd &= 2\r\n\\end{aligned}\r\n$$\r\nNow find $x$ and $y$:\r\n$x$ is the 2nd term: $x = a+d = 5+2 = 7$.\r\n$y$ is the 3rd term: $y = a+2d = 5+2(2) = 9$.\r\n\r\nThus, **$x=7$ and $y=9$**.\r\n\r\n***\r\n\r\n#### Question 14: Sum of a G.P.\r\n**Find the sum of the series: 2 + 4 + 8 + ... upto 8 terms.**\r\n\r\n**Solution:**\r\nThis is a G.P. with $a=2$, $r=2$, and $n=8$.\r\nSince $r > 1$, we use the formula $S_n = \\frac{a(r^n - 1)}{r-1}$.\r\n$$\r\n\\begin{aligned}\r\nS_8 &= \\frac{2(2^8 - 1)}{2-1} \\\\\r\n&= \\frac{2(256 - 1)}{1} \\\\\r\n&= 2(255) \\\\\r\n&= 510\r\n\\end{aligned}\r\n$$\r\nThe sum of the first 8 terms is **510**.\r\n\r\n***\r\n\r\n### Long Questions\r\n\r\n***\r\n\r\n#### Question 15: A.P. and G.P. Relationship\r\n**The product of the first five terms of a geometric series is 243. If the third term of the geometric series is equal to the tenth term of an arithmetic series, find the sum of the first 19 terms of the arithmetic series.**\r\n\r\n**Solution:**\r\n**For the G.P.:**\r\nProduct $P_5 = (a)(ar)(ar^2)(ar^3)(ar^4) = a^5r^{10} = (ar^2)^5$.\r\n$$\r\n\\begin{aligned}\r\n(ar^2)^5 &= 243 \\\\\r\n(ar^2)^5 &= 3^5 \\\\\r\nar^2 &= 3\r\n\\end{aligned}\r\n$$\r\nSo, the third term of the G.P. ($t_3$) is 3.\r\n\r\n**For the A.P.:**\r\nWe are given that the tenth term ($t_{10}$) of the A.P. is equal to the third term of the G.P.\r\n$$\r\nt_{10} = a + (10-1)d = a + 9d = 3\r\n$$\r\n**Sum of the A.P.:**\r\nWe need to find the sum of the first 19 terms ($S_{19}$).\r\n$$\r\n\\begin{aligned}\r\nS_{19} &= \\frac{19}{2}[2a + (19-1)d] \\\\\r\n&= \\frac{19}{2}[2a + 18d] \\\\\r\n&= \\frac{19}{2} \\cdot 2(a + 9d) \\\\\r\n&= 19(a + 9d)\r\n\\end{aligned}\r\n$$\r\nSince we know $a+9d = 3$:\r\n$$\r\nS_{19} = 19(3) = 57\r\n$$\r\nThe sum of the first 19 terms is **57**.\r\n\r\n***\r\n\r\n#### Question 16: Finding Terms from Means\r\n**There are five geometric means between a and b. If the second mean is 2 and the last mean is 16, find the values of a and b.**\r\n\r\n**Solution:**\r\nThe sequence is $a, m_1, m_2, m_3, m_4, m_5, b$.\r\nThis is a G.P. with $n = 5+2 = 7$ terms.\r\n$a$ is the 1st term.\r\n$m_2$ (second mean) is the 3rd term of the sequence: $t_3 = ar^2 = 2$.\r\n$m_5$ (last mean) is the 6th term of the sequence: $t_6 = ar^5 = 16$.\r\n\r\nDivide the two equations:\r\n$$\r\n\\begin{aligned}\r\n\\frac{ar^5}{ar^2} &= \\frac{16}{2} \\\\\r\nr^3 &= 8 \\\\\r\nr &= 2\r\n\\end{aligned}\r\n$$\r\nNow find $a$ using $ar^2 = 2$:\r\n$$\r\n\\begin{aligned}\r\na(2)^2 &= 2 \\\\\r\n4a &= 2 \\\\\r\na &= \\frac{1}{2}\r\n\\end{aligned}\r\n$$\r\nFind $b$, which is the 7th term ($t_7$):\r\n$$\r\nb = ar^6 = \\left(\\frac{1}{2}\\right)(2)^6 = \\frac{64}{2} = 32\r\n$$\r\nThus, **$a = 1/2$ and $b = 32$**.\r\n\r\n***\r\n\r\n#### Question 17: Sum of an A.P. from Sums\r\n**If the sum of the first 9 terms of an arithmetic series is 72 and the sum of the first 17 terms is 289, find the sum of the first 25 terms.**\r\n\r\n**Solution:**\r\nGiven $S_9 = 72$ and $S_{17} = 289$.\r\nFor $S_9$:\r\n$$\r\n\\begin{aligned}\r\n\\frac{9}{2}(2a + 8d) &= 72 \\\\\r\n9(a+4d) &= 72 \\\\\r\na+4d &= 8 \\implies a = 8-4d \\quad \\text{(Eq. 1)}\r\n\\end{aligned}\r\n$$\r\nFor $S_{17}$:\r\n$$\r\n\\begin{aligned}\r\n\\frac{17}{2}(2a + 16d) &= 289 \\\\\r\n17(a+8d) &= 289 \\\\\r\na+8d &= 17 \\quad \\text{(Eq. 2)}\r\n\\end{aligned}\r\n$$\r\nSubstitute Eq. 1 into Eq. 2:\r\n$$\r\n\\begin{aligned}\r\n(8-4d)+8d &= 17 \\\\\r\n8+4d &= 17 \\\\\r\n4d &= 9 \\implies d=9/4\r\n\\end{aligned}\r\n$$\r\nNow find $a$: $a = 8 - 4(9/4) = 8-9 = -1$.\r\n\r\nFind $S_{25}$:\r\n$$\r\n\\begin{aligned}\r\nS_{25} &= \\frac{25}{2}[2(-1) + (25-1)(9/4)] \\\\\r\n&= \\frac{25}{2}[-2 + 24(9/4)] \\\\\r\n&= \\frac{25}{2}[-2 + 6 \\times 9] \\\\\r\n&= \\frac{25}{2}[-2 + 54] = \\frac{25}{2}(52) \\\\\r\n&= 25 \\times 26 = 650\r\n\\end{aligned}\r\n$$\r\nThe sum of the first 25 terms is **650**.\r\n\r\n***\r\n\r\n#### Question 18: Sum of a G.P. from Two Terms\r\n**If the second term and fifth term of the geometric series are 15 and 405 respectively, find the sum of the first 6 terms of the series.**\r\n\r\n**Solution:**\r\nGiven $t_2 = ar = 15$ and $t_5 = ar^4 = 405$.\r\nDivide the two equations:\r\n$$\r\n\\begin{aligned}\r\n\\frac{ar^4}{ar} &= \\frac{405}{15} \\\\\r\nr^3 &= 27 \\\\\r\nr &= 3\r\n\\end{aligned}\r\n$$\r\nFind $a$: $a(3) = 15 \\implies a=5$.\r\nFind $S_6$ (since $r>1$):\r\n$$\r\n\\begin{aligned}\r\nS_6 &= \\frac{5(3^6-1)}{3-1} \\\\\r\n&= \\frac{5(729-1)}{2} \\\\\r\n&= \\frac{5(728)}{2} = 5(364) = 1820\r\n\\end{aligned}\r\n$$\r\nThe sum of the first 6 terms is **1820**.\r\n\r\n***\r\n\r\n#### Question 19: Sum of a G.P. from Two Terms\r\n**Find the sum of the first 9 terms of a geometric series whose third term and seventh term are 20 and 320 respectively.**\r\n\r\n**Solution:**\r\nGiven $t_3 = ar^2 = 20$ and $t_7 = ar^6 = 320$.\r\nDivide the two equations:\r\n$$\r\n\\begin{aligned}\r\n\\frac{ar^6}{ar^2} &= \\frac{320}{20} \\\\\r\nr^4 &= 16 \\\\\r\nr &= 2\r\n\\end{aligned}\r\n$$\r\nFind $a$: $a(2)^2 = 20 \\implies 4a=20 \\implies a=5$.\r\nFind $S_9$:\r\n$$\r\n\\begin{aligned}\r\nS_9 &= \\frac{5(2^9-1)}{2-1} \\\\\r\n&= 5(512-1) \\\\\r\n&= 5(511) = 2555\r\n\\end{aligned}\r\n$$\r\nThe sum of the first 9 terms is **2555**.\r\n\r\n***\r\n\r\n#### Question 20: Inserting Arithmetic Means\r\n**Insert four arithmetic means between 5 and 25.**\r\n\r\n**Solution:**\r\nWe are inserting $n=4$ means between $a=5$ and $b=25$.\r\nFirst, find the common difference $d$:\r\n$$\r\nd = \\frac{b-a}{n+1} = \\frac{25-5}{4+1} = \\frac{20}{5} = 4\r\n$$\r\nThe means are:\r\n* $m_1 = a+d = 5+4 = 9$\r\n* $m_2 = a+2d = 5+2(4) = 13$\r\n* $m_3 = a+3d = 5+3(4) = 17$\r\n* $m_4 = a+4d = 5+4(4) = 21$\r\n\r\nThe four arithmetic means are **9, 13, 17, 21**.\r\n\r\n***\r\n\r\n#### Question 21: Sum of a G.P.\r\n**Find the sum of the first 8 terms of a geometric series whose fourth term is 54 and common ratio is 3.**\r\n\r\n**Solution:**\r\nGiven $t_4 = 54$ and $r=3$.\r\nFirst, find the first term $a$.\r\n$$\r\n\\begin{aligned}\r\nt_4 &= ar^{3} \\\\\r\n54 &= a(3)^3 \\\\\r\n54 &= 27a \\\\\r\na &= 2\r\n\\end{aligned}\r\n$$\r\nNow, find the sum of the first 8 terms ($S_8$):\r\n$$\r\n\\begin{aligned}\r\nS_8 &= \\frac{a(r^8-1)}{r-1} \\\\\r\n&= \\frac{2(3^8-1)}{3-1} \\\\\r\n&= \\frac{2(6561-1)}{2} \\\\\r\n&= 6560\r\n\\end{aligned}\r\n$$\r\nThe sum is **6560**.\r\n\r\n***\r\n\r\n#### Question 22: Sum and Product of an A.P.\r\n**The sum of three numbers in an arithmetic progression is 51 and the sum of their squares is 939. Find the numbers.**\r\n\r\n**Solution:**\r\nLet the three terms be $a-d, a, a+d$.\r\nFrom the sum:\r\n$$\r\n\\begin{aligned}\r\n(a-d) + a + (a+d) &= 51 \\\\\r\n3a &= 51 \\\\\r\na &= 17\r\n\\end{aligned}\r\n$$\r\nFrom the sum of squares:\r\n$$\r\n\\begin{aligned}\r\n(a-d)^2 + a^2 + (a+d)^2 &= 939 \\\\\r\n(17-d)^2 + 17^2 + (17+d)^2 &= 939 \\\\\r\n(289 - 34d + d^2) + 289 + (289 + 34d + d^2) &= 939 \\\\\r\n867 + 2d^2 &= 939 \\\\\r\n2d^2 &= 72 \\\\\r\nd^2 &= 36 \\\\\r\nd &= \\pm 6\r\n\\end{aligned}\r\n$$\r\nIf $d=6$, the numbers are $11, 17, 23$.\r\nIf $d=-6$, the numbers are $23, 17, 11$.\r\nThe numbers are **11, 17, and 23**.\r\n\r\n***\r\n\r\n#### Question 23: Sum and Product of an A.P.\r\n**Three terms in an arithmetic progression have sum 21 and product 315. Find the terms.**\r\n\r\n**Solution:**\r\nLet the three terms be $a-d, a, a+d$.\r\nFrom the sum:\r\n$$\r\n(a-d) + a + (a+d) = 21 \\implies 3a = 21 \\implies a=7\r\n$$\r\nFrom the product:\r\n$$\r\n\\begin{aligned}\r\n(a-d)(a)(a+d) &= 315 \\\\\r\n(7-d)(7)(7+d) &= 315 \\\\\r\n(49-d^2)(7) &= 315 \\\\\r\n49-d^2 &= 45 \\\\\r\nd^2 &= 4 \\\\\r\nd &= \\pm 2\r\n\\end{aligned}\r\n$$\r\nIf $d=2$, the numbers are $5, 7, 9$.\r\nIf $d=-2$, the numbers are $9, 7, 5$.\r\nThe terms are **5, 7, and 9**.\r\n\r\n***\r\n\r\n#### Question 24: Sum of an A.P. from Two Terms\r\n**The ninth and nineteenth terms of an arithmetic series are 40 and 60 respectively. Find the sum of the first 25 terms of the series.**\r\n\r\n**Solution:**\r\nGiven: $t_9 = a+8d = 40$ and $t_{19} = a+18d = 60$.\r\nSubtract the first equation from the second:\r\n$$\r\n(a+18d) - (a+8d) = 60-40 \\implies 10d=20 \\implies d=2\r\n$$\r\nFind $a$: $a+8(2)=40 \\implies a+16=40 \\implies a=24$.\r\nFind $S_{25}$:\r\n$$\r\n\\begin{aligned}\r\nS_{25} &= \\frac{25}{2}[2(24) + (25-1)2] \\\\\r\n&= \\frac{25}{2}[48 + 24(2)] \\\\\r\n&= \\frac{25}{2}[48+48] = \\frac{25}{2}(96) \\\\\r\n&= 25(48) = 1200\r\n\\end{aligned}\r\n$$\r\nThe sum is **1200**.\r\n\r\n***\r\n\r\n#### Question 25: A.P. Word Problem\r\n**In an arithmetic sequence, the sixth term is equal to three times the fourth term and the sum of the first three terms is -12. Find the sum of the first ten terms.**\r\n\r\n**Solution:**\r\nFrom the first condition: $t_6 = 3 \\times t_4$.\r\n$$\r\n\\begin{aligned}\r\na+5d &= 3(a+3d) \\\\\r\na+5d &= 3a+9d \\\\\r\n-2a &= 4d \\\\\r\na &= -2d \\quad \\text{(Eq. 1)}\r\n\\end{aligned}\r\n$$\r\nFrom the second condition, let the first three terms be $a-d, a, a+d$.\r\nTheir sum is $(a-d)+a+(a+d) = 3a = -12 \\implies a=-4$.\r\n\r\nSubstitute $a=-4$ into Eq. 1:\r\n$-4 = -2d \\implies d=2$.\r\n\r\nFind $S_{10}$:\r\n$$\r\n\\begin{aligned}\r\nS_{10} &= \\frac{10}{2}[2(-4) + (10-1)2] \\\\\r\n&= 5[-8 + 9(2)] \\\\\r\n&= 5[-8+18] = 5(10) = 50\r\n\\end{aligned}\r\n$$\r\nThe sum is **50**.\r\n\r\n***\r\n\r\n#### Question 26: A.P. Word Problem\r\n**The sum of the first eight terms of an arithmetic series is 180 and its fifth term is five times the first term. Find the first term and common difference.**\r\n\r\n**Solution:**\r\nFrom the first condition: $t_5 = 5a$.\r\n$$\r\na+4d = 5a \\implies 4d=4a \\implies d=a \\quad \\text{(Eq. 1)}\r\n$$\r\nFrom the second condition: $S_8 = 180$.\r\n$$\r\n\\begin{aligned}\r\n\\frac{8}{2}[2a+(8-1)d] &= 180 \\\\\r\n4[2a+7d] &= 180 \\\\\r\n2a+7d &= 45\r\n\\end{aligned}\r\n$$\r\nSubstitute $d=a$ into this equation:\r\n$$\r\n2a+7a = 45 \\implies 9a = 45 \\implies a=5\r\n$$\r\nSince $d=a$, then $d=5$.\r\nThe first term is **5** and the common difference is **5**.\r\n\r\n***\r\n\r\n#### Question 27: Sum of a G.P. from Sums\r\n**In a geometric series, the sum of the first three terms is 1 and the sum of its first six terms is 28. Find the common ratio and first term.**\r\n\r\n**Solution:**\r\nGiven $S_3 = \\frac{a(r^3-1)}{r-1} = 1$ and $S_6 = \\frac{a(r^6-1)}{r-1} = 28$.\r\nWe can rewrite $S_6$:\r\n$$\r\nS_6 = \\frac{a(r^3-1)(r^3+1)}{r-1} = \\left[\\frac{a(r^3-1)}{r-1}\\right](r^3+1) = S_3(r^3+1)\r\n$$\r\nNow substitute the known values:\r\n$$\r\n\\begin{aligned}\r\n28 &= 1(r^3+1) \\\\\r\n27 &= r^3 \\\\\r\nr &= 3\r\n\\end{aligned}\r\n$$\r\nNow find $a$ using the equation for $S_3$:\r\n$$\r\n\\begin{aligned}\r\n\\frac{a(3^3-1)}{3-1} &= 1 \\\\\r\n\\frac{a(26)}{2} &= 1 \\\\\r\n13a &= 1 \\\\\r\na &= \\frac{1}{13}\r\n\\end{aligned}\r\n$$\r\nThe common ratio is **3** and the first term is **1/13**.', 1, '2025-07-25 01:11:38', '2025-09-18 07:21:49'),
(141, 13104, '# Grade 10 Optional Mathematics: Matrix Study Guide\r\n## Unit: Algebra | Chapter: Matrix | Complete Solutions\r\n\r\n---\r\n\r\n## 🎯 Chapter Introduction\r\n\r\nWelcome to the comprehensive study guide for the **Matrix** chapter. This note is designed to make complex concepts easy to understand, with clear definitions, essential formulas, and detailed, step-by-step solutions to a wide range of problems. By the end of this guide, you will have mastered matrix operations, determinants, inverses, and solving systems of equations using Cramer\'s Rule. 📈\r\n\r\n### Key Learning Areas\r\n1.  **Matrix Fundamentals:** Definitions, order, and types of matrices.\r\n2.  **Matrix Operations:** Multiplication of matrices and its conditions.\r\n3.  **Determinants:** How to calculate the determinant of a 2x2 matrix.\r\n4.  **Inverse of a Matrix:** Finding the adjoint and inverse of a 2x2 matrix.\r\n5.  **Cramer\'s Rule:** A method for solving systems of linear equations.\r\n\r\n---\r\n\r\n## 🔑 Key Concepts & Formulas\r\n\r\nBefore we solve the problems, let\'s understand the main tools we\'ll be using.\r\n\r\n### 1. Core Definitions\r\n\r\n* **Matrix:** A rectangular array of numbers arranged in rows and columns, enclosed in brackets `[]` or parentheses `()`.\r\n* **Order of a Matrix:** A matrix with `m` rows and `n` columns has an order of `m x n`.\r\n* **Identity Matrix (I):** A square matrix with `1`s on the main diagonal and `0`s elsewhere. For a 2x2 matrix: $I = \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}$. Multiplying any matrix by `I` leaves it unchanged (i.e., `AI = IA = A`).\r\n* **Null Matrix (O):** A matrix where all elements are zero.\r\n* **Transpose of a Matrix ($A^T$):** The matrix obtained by interchanging the rows and columns of matrix A.\r\n* **Inverse Matrix ($A^{-1}$):** If A and B are two square matrices of the same order such that their product is the identity matrix ($AB = BA = I$), then B is the inverse of A, and A is the inverse of B.\r\n\r\n---\r\n\r\n### 2. Matrix Multiplication\r\n\r\n* **Condition:** Two matrices A (order `m x n`) and B (order `p x q`) can be multiplied to get the product AB only if the **number of columns in A is equal to the number of rows in B** (i.e., `n = p`).\r\n    * The resulting matrix AB will have an order of `m x q`.\r\n\r\n---\r\n\r\n### 3. Determinant of a 2x2 Matrix\r\n\r\nThe determinant of a matrix is a special number that can be calculated from its elements. For a matrix $A = \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}$, the determinant is denoted as `|A|` or `det(A)`.\r\n\r\n* **Formula:**\r\n    $$|A| = ad - bc$$\r\n\r\n* **Singular Matrix:** A matrix whose determinant is zero (`|A| = 0`). A singular matrix does **not** have an inverse. This is the necessary condition for an inverse to be undefined.\r\n\r\n---\r\n\r\n### 4. Inverse of a 2x2 Matrix\r\n\r\n* **Condition for Existence:** The inverse of a matrix A exists **only if its determinant is not zero** (i.e., $|A| \\neq 0$).\r\n\r\n* **Steps to Find the Inverse:**\r\n    1.  **Find the Determinant:** Calculate $|A| = ad - bc$. If it\'s zero, stop. The inverse does not exist.\r\n    2.  **Find the Adjoint:** For a matrix $A = \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}$, the adjoint of A, denoted `adj(A)`, is found by swapping the elements of the main diagonal and changing the signs of the other two elements:\r\n        $$adj(A) = \\begin{bmatrix} d & -b \\\\ -c & a \\end{bmatrix}$$\r\n    3.  **Calculate the Inverse:** Use the formula:\r\n        $$A^{-1} = \\frac{1}{|A|} \\text{adj}(A)$$\r\n\r\n---\r\n\r\n### 5. Cramer\'s Rule\r\n\r\nCramer\'s Rule is a method for solving systems of linear equations using determinants.\r\nFor a system of two linear equations:\r\n$$\r\n\\begin{aligned}\r\na_1x + b_1y &= c_1 \\\\\r\na_2x + b_2y &= c_2\r\n\\end{aligned}\r\n$$\r\n**Step 1: Define the Determinants.**\r\n* **D (Coefficient Determinant):** The determinant of the coefficient matrix.\r\n    $$D = \\begin{vmatrix} a_1 & b_1 \\\\ a_2 & b_2 \\end{vmatrix}$$\r\n* **$D_x$:** Replace the x-column in D with the constant terms.\r\n    $$D_x = \\begin{vmatrix} c_1 & b_1 \\\\ c_2 & b_2 \\end{vmatrix}$$\r\n* **$D_y$:** Replace the y-column in D with the constant terms.\r\n    $$D_y = \\begin{vmatrix} a_1 & c_1 \\\\ a_2 & c_2 \\end{vmatrix}$$\r\n\r\n\r\n\r\n**Step 2: Solve for x and y.**\r\n$$\r\nx = \\frac{D_x}{D} \\quad \\text{and} \\quad y = \\frac{D_y}{D}\r\n$$\r\n\r\n* **Condition for a Unique Solution:** A unique solution exists only if the determinant of the coefficient matrix is non-zero ($D \\neq 0$).\r\n\r\n---\r\n\r\n## 📚 Exercise Solutions\r\n\r\n### Short Questions\r\n\r\n***\r\n\r\n#### Question 1: Find an Unknown Matrix\r\n**If $\\begin{bmatrix} 4 \\\\ 1 \\end{bmatrix} \\cdot C = \\begin{bmatrix} -8 & 12 \\\\ -2 & 3 \\end{bmatrix}$, find the matrix C.**\r\n\r\n**Solution:**\r\nThe first matrix is $2 \\times 1$. The resulting matrix is $2 \\times 2$.\r\nFor the multiplication to be possible, matrix C must have an order of $1 \\times 2$.\r\nLet $C = \\begin{bmatrix} a & b \\end{bmatrix}$.\r\n\r\n$$\r\n\\begin{aligned}\r\n\\begin{bmatrix} 4 \\\\ 1 \\end{bmatrix} \\begin{bmatrix} a & b \\end{bmatrix} &= \\begin{bmatrix} -8 & 12 \\\\ -2 & 3 \\end{bmatrix} \\\\\r\n\\begin{bmatrix} 4 \\cdot a & 4 \\cdot b \\\\ 1 \\cdot a & 1 \\cdot b \\end{bmatrix} &= \\begin{bmatrix} -8 & 12 \\\\ -2 & 3 \\end{bmatrix} \\\\\r\n\\begin{bmatrix} 4a & 4b \\\\ a & b \\end{bmatrix} &= \\begin{bmatrix} -8 & 12 \\\\ -2 & 3 \\end{bmatrix}\r\n\\end{aligned}\r\n$$\r\nEquating the corresponding elements:\r\nFrom the first column: $a = -2$.\r\nFrom the second column: $b = 3$.\r\n\r\nThus, **$C = \\begin{bmatrix} -2 & 3 \\end{bmatrix}$**.\r\n\r\n***\r\n\r\n#### Question 2: Find an Unknown Matrix\r\n**If $\\begin{bmatrix} 4 \\\\ 3 \\end{bmatrix} \\cdot p = \\begin{bmatrix} 8 & -4 \\\\ 6 & -3 \\end{bmatrix}$, find the matrix p.**\r\n\r\n**Solution:**\r\nLet the matrix $p$ be a $1 \\times 2$ matrix, $p = \\begin{bmatrix} a & b \\end{bmatrix}$.\r\n\r\n$$\r\n\\begin{aligned}\r\n\\begin{bmatrix} 4 \\\\ 3 \\end{bmatrix} \\begin{bmatrix} a & b \\end{bmatrix} &= \\begin{bmatrix} 8 & -4 \\\\ 6 & -3 \\end{bmatrix} \\\\\r\n\\begin{bmatrix} 4a & 4b \\\\ 3a & 3b \\end{bmatrix} &= \\begin{bmatrix} 8 & -4 \\\\ 6 & -3 \\end{bmatrix}\r\n\\end{aligned}\r\n$$\r\nEquating the corresponding elements:\r\nFrom the first column: $3a = 6 \\implies a=2$.\r\nFrom the second column: $3b = -3 \\implies b=-1$.\r\n\r\nThus, **$p = \\begin{bmatrix} 2 & -1 \\end{bmatrix}$**.\r\n\r\n***\r\n\r\n#### Question 3: Matrix Multiplication\r\n**If $A = \\begin{bmatrix} 1 & 2 & -3 \\\\ -2 & 0 & 5 \\end{bmatrix}$ and $B = \\begin{bmatrix} 2 & 3 \\\\ -4 & 0 \\\\ 5 & -1 \\end{bmatrix}$, find AB.**\r\n\r\n**Solution:**\r\nMatrix A is $2 \\times 3$ and B is $3 \\times 2$. The product AB will be a $2 \\times 2$ matrix.\r\n$$\r\n\\begin{aligned}\r\nAB &= \\begin{bmatrix} 1 & 2 & -3 \\\\ -2 & 0 & 5 \\end{bmatrix} \\begin{bmatrix} 2 & 3 \\\\ -4 & 0 \\\\ 5 & -1 \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} (1)(2) + (2)(-4) + (-3)(5) & (1)(3) + (2)(0) + (-3)(-1) \\\\ (-2)(2) + (0)(-4) + (5)(5) & (-2)(3) + (0)(0) + (5)(-1) \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} 2 - 8 - 15 & 3 + 0 + 3 \\\\ -4 + 0 + 25 & -6 + 0 - 5 \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} -21 & 6 \\\\ 21 & -11 \\end{bmatrix}\r\n\\end{aligned}\r\n$$\r\nThus, **$AB = \\begin{bmatrix} -21 & 6 \\\\ 21 & -11 \\end{bmatrix}$**.\r\n\r\n***\r\n\r\n#### Question 4: Singular Matrix\r\n**If the inverse of the matrix $\\begin{bmatrix} x & 3 \\\\ 4 & 2 \\end{bmatrix}$ cannot be defined, find the value of x.**\r\n\r\n**Solution:**\r\nAn inverse of a matrix cannot be defined if the matrix is singular, which means its determinant is 0.\r\nLet $A = \\begin{bmatrix} x & 3 \\\\ 4 & 2 \\end{bmatrix}$.\r\n$$\r\n\\begin{aligned}\r\n|A| &= 0 \\\\\r\n(x)(2) - (3)(4) &= 0 \\\\\r\n2x - 12 &= 0 \\\\\r\n2x &= 12 \\\\\r\nx &= 6\r\n\\end{aligned}\r\n$$\r\nThus, **$x=6$**.\r\n\r\n***\r\n\r\n#### Question 5: Find an Inverse Matrix\r\n**If matrix $A = \\begin{bmatrix} 5 & 3 \\\\ 7 & 4 \\end{bmatrix}$, find $A^{-1}$.**\r\n\r\n**Solution:**\r\n**Step 1: Find the determinant.**\r\n$$\r\n|A| = (5)(4) - (3)(7) = 20 - 21 = -1\r\n$$\r\nSince $|A| \\neq 0$, the inverse exists.\r\n\r\n**Step 2: Find the adjoint.**\r\n$$\r\nadj(A) = \\begin{bmatrix} 4 & -3 \\\\ -7 & 5 \\end{bmatrix}\r\n$$\r\n**Step 3: Calculate the inverse.**\r\n$$\r\nA^{-1} = \\frac{1}{|A|} \\text{adj}(A) = \\frac{1}{-1} \\begin{bmatrix} 4 & -3 \\\\ -7 & 5 \\end{bmatrix} = \\begin{bmatrix} -4 & 3 \\\\ 7 & -5 \\end{bmatrix}\r\n$$\r\nThus, **$A^{-1} = \\begin{bmatrix} -4 & 3 \\\\ 7 & -5 \\end{bmatrix}$**.\r\n\r\n***\r\n\r\n#### Question 6: Finding a Value from an Inverse\r\n**If the inverse of the matrix $\\begin{bmatrix} m & 2 \\\\ 7 & 3 \\end{bmatrix}$ is the matrix $\\begin{bmatrix} 3 & -2 \\\\ -7 & m \\end{bmatrix}$, find the value of m.**\r\n\r\n**Solution:**\r\nLet $A = \\begin{bmatrix} m & 2 \\\\ 7 & 3 \\end{bmatrix}$ and $A^{-1} = \\begin{bmatrix} 3 & -2 \\\\ -7 & m \\end{bmatrix}$.\r\nBy the definition of an inverse, $A \\cdot A^{-1} = I$.\r\n$$\r\n\\begin{aligned}\r\n\\begin{bmatrix} m & 2 \\\\ 7 & 3 \\end{bmatrix} \\begin{bmatrix} 3 & -2 \\\\ -7 & m \\end{bmatrix} &= \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix} \\\\\r\n\\begin{bmatrix} 3m - 14 & -2m + 2m \\\\ 21 - 21 & -14 + 3m \\end{bmatrix} &= \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix} \\\\\r\n\\begin{bmatrix} 3m - 14 & 0 \\\\ 0 & 3m - 14 \\end{bmatrix} &= \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}\r\n\\end{aligned}\r\n$$\r\nEquating the top-left elements:\r\n$$\r\n\\begin{aligned}\r\n3m - 14 &= 1 \\\\\r\n3m &= 15 \\\\\r\nm &= 5\r\n\\end{aligned}\r\n$$\r\nThus, **$m=5$**.\r\n\r\n***\r\n\r\n#### Question 7: Matrix and its Transpose\r\n**If matrix $A = \\begin{bmatrix} 3 & 1 \\\\ -1 & 2 \\end{bmatrix}$, find $A \\cdot A^T$.**\r\n\r\n**Solution:**\r\nFirst, find the transpose of A, $A^T$.\r\n$$\r\nA^T = \\begin{bmatrix} 3 & -1 \\\\ 1 & 2 \\end{bmatrix}\r\n$$\r\nNow, multiply A by $A^T$.\r\n$$\r\n\\begin{aligned}\r\nA \\cdot A^T &= \\begin{bmatrix} 3 & 1 \\\\ -1 & 2 \\end{bmatrix} \\begin{bmatrix} 3 & -1 \\\\ 1 & 2 \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} (3)(3) + (1)(1) & (3)(-1) + (1)(2) \\\\ (-1)(3) + (2)(1) & (-1)(-1) + (2)(2) \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} 9+1 & -3+2 \\\\ -3+2 & 1+4 \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} 10 & -1 \\\\ -1 & 5 \\end{bmatrix}\r\n\\end{aligned}\r\n$$\r\nThus, **$A \\cdot A^T = \\begin{bmatrix} 10 & -1 \\\\ -1 & 5 \\end{bmatrix}$**.\r\n\r\n***\r\n\r\n#### Question 8: Find an Unknown from a Matrix Equation\r\n**If $A = \\begin{bmatrix} 2 & 0 \\\\ -3 & x \\end{bmatrix}$, $B = \\begin{bmatrix} 4 & 0 \\\\ -9 & 1 \\end{bmatrix}$ and $A^2 = B$, find the value of x.**\r\n\r\n**Solution:**\r\nFirst, calculate $A^2$.\r\n$$\r\n\\begin{aligned}\r\nA^2 &= A \\cdot A = \\begin{bmatrix} 2 & 0 \\\\ -3 & x \\end{bmatrix} \\begin{bmatrix} 2 & 0 \\\\ -3 & x \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} (2)(2) + (0)(-3) & (2)(0) + (0)(x) \\\\ (-3)(2) + (x)(-3) & (-3)(0) + (x)(x) \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} 4 & 0 \\\\ -6 - 3x & x^2 \\end{bmatrix}\r\n\\end{aligned}\r\n$$\r\nNow, set $A^2 = B$.\r\n$$\r\n\\begin{bmatrix} 4 & 0 \\\\ -6 - 3x & x^2 \\end{bmatrix} = \\begin{bmatrix} 4 & 0 \\\\ -9 & 1 \\end{bmatrix}\r\n$$\r\nEquating the bottom-right elements:\r\n$$x^2 = 1 \\implies x = \\pm 1$$\r\nEquating the bottom-left elements:\r\n$$-6 - 3x = -9 \\implies -3x = -3 \\implies x = 1$$\r\nBoth conditions must be true, so the only valid solution is **$x=1$**.\r\n\r\n***\r\n\r\n#### Question 9: Finding a Value from Determinant\r\n**If $M = \\begin{bmatrix} 5 & 3 \\\\ 6 & K \\end{bmatrix}$ and $|M| = 27$, find the value of K.**\r\n\r\n**Solution:**\r\nCalculate the determinant of M and set it equal to 27.\r\n$$\r\n\\begin{aligned}\r\n|M| &= (5)(K) - (3)(6) \\\\\r\n27 &= 5K - 18 \\\\\r\n45 &= 5K \\\\\r\nK &= 9\r\n\\end{aligned}\r\n$$\r\nThus, **$K=9$**.\r\n\r\n***\r\n\r\n#### Question 10: Prove AB is a Null Matrix\r\n**If $A = \\begin{bmatrix} 1 & -1 \\\\ 3 & -3 \\end{bmatrix}$ and $B = \\begin{bmatrix} 2 & -5 \\\\ 2 & -5 \\end{bmatrix}$, show that AB is a null matrix.**\r\n\r\n**Solution:**\r\nMultiply the matrices A and B.\r\n$$\r\n\\begin{aligned}\r\nAB &= \\begin{bmatrix} 1 & -1 \\\\ 3 & -3 \\end{bmatrix} \\begin{bmatrix} 2 & -5 \\\\ 2 & -5 \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} (1)(2) + (-1)(2) & (1)(-5) + (-1)(-5) \\\\ (3)(2) + (-3)(2) & (3)(-5) + (-3)(-5) \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} 2-2 & -5+5 \\\\ 6-6 & -15+15 \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} 0 & 0 \\\\ 0 & 0 \\end{bmatrix}\r\n\\end{aligned}\r\n$$\r\nSince the resulting matrix contains only zeros, **AB is a null matrix**.\r\n\r\n***\r\n\r\n#### Question 11: Finding a Matrix from its Inverse\r\n**If the inverse of a matrix A is $A^{-1} = \\begin{bmatrix} 1 & 3 \\\\ 2 & 4 \\end{bmatrix}$, find the matrix A.**\r\n\r\n**Solution:**\r\nThe inverse of an inverse matrix is the original matrix, i.e., $(A^{-1})^{-1} = A$. So, we find the inverse of the given $A^{-1}$.\r\nLet $M = A^{-1} = \\begin{bmatrix} 1 & 3 \\\\ 2 & 4 \\end{bmatrix}$.\r\n\r\n**Step 1: Find the determinant of M.**\r\n$$\r\n|M| = (1)(4) - (3)(2) = 4 - 6 = -2\r\n$$\r\n**Step 2: Find the adjoint of M.**\r\n$$\r\nadj(M) = \\begin{bmatrix} 4 & -3 \\\\ -2 & 1 \\end{bmatrix}\r\n$$\r\n**Step 3: Calculate the inverse of M.**\r\n$$\r\nA = M^{-1} = \\frac{1}{|M|}adj(M) = \\frac{1}{-2} \\begin{bmatrix} 4 & -3 \\\\ -2 & 1 \\end{bmatrix} = \\begin{bmatrix} -2 & 3/2 \\\\ 1 & -1/2 \\end{bmatrix}\r\n$$\r\nThus, **$A = \\begin{bmatrix} -2 & 3/2 \\\\ 1 & -1/2 \\end{bmatrix}$**.\r\n\r\n***\r\n\r\n#### Question 12: Prove AB is an Identity Matrix\r\n**If $A = \\begin{bmatrix} 3 & 4 \\\\ 2 & 3 \\end{bmatrix}$ and $B = \\begin{bmatrix} 3 & -4 \\\\ -2 & 3 \\end{bmatrix}$, prove that AB is an identity matrix.**\r\n\r\n**Solution:**\r\nMultiply the matrices A and B.\r\n$$\r\n\\begin{aligned}\r\nAB &= \\begin{bmatrix} 3 & 4 \\\\ 2 & 3 \\end{bmatrix} \\begin{bmatrix} 3 & -4 \\\\ -2 & 3 \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} (3)(3) + (4)(-2) & (3)(-4) + (4)(3) \\\\ (2)(3) + (3)(-2) & (2)(-4) + (3)(3) \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} 9-8 & -12+12 \\\\ 6-6 & -8+9 \\end{bmatrix} \\\\\r\n&= \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}\r\n\\end{aligned}\r\n$$\r\nSince the result is $\\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}$, **AB is an identity matrix**.\r\n\r\n***\r\n\r\n#### Question 13: Find Unknowns in a Matrix Equation\r\n**If $\\begin{pmatrix} -2 & 0 \\\\ 0 & -5 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} -2 \\\\ -10 \\end{pmatrix}$, find the values of x and y.**\r\n\r\n**Solution:**\r\nPerform the matrix multiplication on the left side:\r\n$$\r\n\\begin{pmatrix} (-2)(x) + (0)(y) \\\\ (0)(x) + (-5)(y) \\end{pmatrix} = \\begin{pmatrix} -2 \\\\ -10 \\end{pmatrix} \\\\\r\n\\begin{pmatrix} -2x \\\\ -5y \\end{pmatrix} = \\begin{pmatrix} -2 \\\\ -10 \\end{pmatrix}\r\n$$\r\nEquating the corresponding elements:\r\n$$-2x = -2 \\implies x = 1$$\r\n$$-5y = -10 \\implies y = 2$$\r\nThus, **$x=1$ and $y=2$**.\r\n\r\n***\r\n\r\n#### Question 14: Find Unknowns in a Matrix Equation\r\n**If $\\begin{pmatrix} x + 1 \\\\ y - 2 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix} \\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix}$, find the values of x and y.**\r\n\r\n**Solution:**\r\nFirst, perform the matrix multiplication on the right side:\r\n$$\r\n\\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix} \\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix} = \\begin{pmatrix} (1)(2) + (0)(3) \\\\ (0)(2) + (-1)(3) \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ -3 \\end{pmatrix}\r\n$$\r\nNow, equate the matrices:\r\n$$\r\n\\begin{pmatrix} x + 1 \\\\ y - 2 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ -3 \\end{pmatrix}\r\n$$\r\nEquating corresponding elements:\r\n$$x+1 = 2 \\implies x=1$$\r\n$$y-2 = -3 \\implies y = -1$$\r\nThus, **$x=1$ and $y=-1$**.\r\n\r\n***\r\n\r\n#### Question 15: Find a Value from Determinant\r\n**If the determinant of the matrix $M = \\begin{bmatrix} -5 & -8 \\\\ p & 12 \\end{bmatrix}$ is 20, find the value of p.**\r\n\r\n**Solution:**\r\nGiven $|M|=20$.\r\n$$\r\n\\begin{aligned}\r\n|M| &= (-5)(12) - (-8)(p) \\\\\r\n20 &= -60 + 8p \\\\\r\n80 &= 8p \\\\\r\np &= 10\r\n\\end{aligned}\r\n$$\r\nThus, **$p=10$**.\r\n\r\n---\r\n\r\n### Solving Systems of Linear Equations\r\n\r\n***\r\n\r\n#### Question 16: Solve using Cramer\'s Rule\r\n**Solve: $2x - y = 3$ and $y + 3x = 7$.**\r\n\r\n**Solution:**\r\nFirst, arrange the equations in standard form:\r\n$$\r\n\\begin{aligned}\r\n2x - y &= 3 \\\\\r\n3x + y &= 7\r\n\\end{aligned}\r\n$$\r\nNow, find the determinants D, $D_x$, and $D_y$.\r\n$$\r\n\\begin{aligned}\r\nD &= \\begin{vmatrix} 2 & -1 \\\\ 3 & 1 \\end{vmatrix} = (2)(1) - (-1)(3) = 2 + 3 = 5 \\\\\r\nD_x &= \\begin{vmatrix} 3 & -1 \\\\ 7 & 1 \\end{vmatrix} = (3)(1) - (-1)(7) = 3 + 7 = 10 \\\\\r\nD_y &= \\begin{vmatrix} 2 & 3 \\\\ 3 & 7 \\end{vmatrix} = (2)(7) - (3)(3) = 14 - 9 = 5\r\n\\end{aligned}\r\n$$\r\nNow, solve for x and y:\r\n$$\r\nx = \\frac{D_x}{D} = \\frac{10}{5} = 2\r\n$$\r\n$$\r\ny = \\frac{D_y}{D} = \\frac{5}{5} = 1\r\n$$\r\nThe solution is **$x=2, y=1$**.\r\n\r\n***\r\n\r\n#### Question 17: Solve using Cramer\'s Rule\r\n**Solve: $y = 2x$ and $x - \\frac{3}{2}y + 1 = 0$.**\r\n\r\n**Solution:**\r\nArrange the equations in standard form:\r\n$$\r\n\\begin{aligned}\r\n2x - y &= 0 \\\\\r\nx - \\frac{3}{2}y &= -1 \\implies 2x - 3y = -2\r\n\\end{aligned}\r\n$$\r\nFind the determinants:\r\n$$\r\n\\begin{aligned}\r\nD &= \\begin{vmatrix} 2 & -1 \\\\ 2 & -3 \\end{vmatrix} = (2)(-3) - (-1)(2) = -6 + 2 = -4 \\\\\r\nD_x &= \\begin{vmatrix} 0 & -1 \\\\ -2 & -3 \\end{vmatrix} = (0)(-3) - (-1)(-2) = 0 - 2 = -2 \\\\\r\nD_y &= \\begin{vmatrix} 2 & 0 \\\\ 2 & -2 \\end{vmatrix} = (2)(-2) - (0)(2) = -4 - 0 = -4\r\n\\end{aligned}\r\n$$\r\nSolve for x and y:\r\n$$\r\nx = \\frac{D_x}{D} = \\frac{-2}{-4} = \\frac{1}{2}\r\n$$\r\n$$\r\ny = \\frac{D_y}{D} = \\frac{-4}{-4} = 1\r\n$$\r\nThe solution is **$x = 1/2, y = 1$**.\r\n\r\n***\r\n\r\n#### Question 18: Solve using Cramer\'s Rule\r\n**Solve: $\\frac{4}{x} + \\frac{5}{y} = 28$ and $\\frac{7}{x} + \\frac{3}{y} = 67$.**\r\n\r\n**Solution:**\r\nLet $u = \\frac{1}{x}$ and $v = \\frac{1}{y}$. The equations become:\r\n$$\r\n\\begin{aligned}\r\n4u + 5v &= 28 \\\\\r\n7u + 3v &= 67\r\n\\end{aligned}\r\n$$\r\nFind the determinants D, $D_u$, and $D_v$.\r\n$$\r\n\\begin{aligned}\r\nD &= \\begin{vmatrix} 4 & 5 \\\\ 7 & 3 \\end{vmatrix} = (4)(3) - (5)(7) = 12 - 35 = -23 \\\\\r\nD_u &= \\begin{vmatrix} 28 & 5 \\\\ 67 & 3 \\end{vmatrix} = (28)(3) - (5)(67) = 84 - 335 = -251 \\\\\r\nD_v &= \\begin{vmatrix} 4 & 28 \\\\ 7 & 67 \\end{vmatrix} = (4)(67) - (28)(7) = 268 - 196 = 72\r\n\\end{aligned}\r\n$$\r\nSolve for u and v:\r\n$$\r\nu = \\frac{D_u}{D} = \\frac{-251}{-23} = 11 \\quad \\implies \\frac{1}{x} = 11 \\implies x = \\frac{1}{11}\r\n$$\r\n$$\r\nv = \\frac{D_v}{D} = \\frac{72}{-23} \\quad \\implies \\frac{1}{y} = -\\frac{72}{23} \\implies y = -\\frac{23}{72}\r\n$$\r\nThe solution is **$x=\\frac{1}{11}, y=-\\frac{23}{72}$**.\r\n\r\n***\r\n\r\n#### Question 19: Solve using Cramer\'s Rule\r\n**Solve: $3y + 4x = 2xy$ and $18y - 4x = 5xy$.**\r\n\r\n**Solution:**\r\nDivide both equations by $xy$ to get them in linear form.\r\nEquation 1:\r\n$$\r\n\\frac{3y}{xy} + \\frac{4x}{xy} = 2 \\implies \\frac{3}{x} + \\frac{4}{y} = 2\r\n$$\r\nEquation 2:\r\n$$\r\n\\frac{18y}{xy} - \\frac{4x}{xy} = 5 \\implies \\frac{18}{x} - \\frac{4}{y} = 5\r\n$$\r\nLet $u = \\frac{1}{x}$ and $v = \\frac{1}{y}$. The system is:\r\n$$\r\n\\begin{aligned}\r\n3u + 4v &= 2 \\\\\r\n18u - 4v &= 5\r\n\\end{aligned}\r\n$$\r\nFind the determinants:\r\n$$\r\n\\begin{aligned}\r\nD &= \\begin{vmatrix} 3 & 4 \\\\ 18 & -4 \\end{vmatrix} = (3)(-4) - (4)(18) = -12 - 72 = -84 \\\\\r\nD_u &= \\begin{vmatrix} 2 & 4 \\\\ 5 & -4 \\end{vmatrix} = (2)(-4) - (4)(5) = -8 - 20 = -28 \\\\\r\nD_v &= \\begin{vmatrix} 3 & 2 \\\\ 18 & 5 \\end{vmatrix} = (3)(5) - (2)(18) = 15 - 36 = -21\r\n\\end{aligned}\r\n$$\r\nSolve for u and v:\r\n$$\r\nu = \\frac{D_u}{D} = \\frac{-28}{-84} = \\frac{1}{3} \\quad \\implies \\frac{1}{x} = \\frac{1}{3} \\implies x = 3\r\n$$\r\n$$\r\nv = \\frac{D_v}{D} = \\frac{-21}{-84} = \\frac{1}{4} \\quad \\implies \\frac{1}{y} = \\frac{1}{4} \\implies y = 4\r\n$$\r\nThe solution is **$x=3, y=4$**.\r\n\r\n***\r\n\r\n#### Question 20: Solve using Cramer\'s Rule\r\n**Solve: $4(x - 1) + 5(y + 2) = 10$ and $5(x - 1) - 3(y + 2) + 6 = 0$.**\r\n\r\n**Solution:**\r\nFirst, simplify both equations.\r\nEquation 1:\r\n$$\r\n\\begin{aligned}\r\n4x - 4 + 5y + 10 &= 10 \\\\\r\n4x + 5y + 6 &= 10 \\\\\r\n4x + 5y &= 4\r\n\\end{aligned}\r\n$$\r\nEquation 2:\r\n$$\r\n\\begin{aligned}\r\n5x - 5 - 3y - 6 + 6 &= 0 \\\\\r\n5x - 3y - 5 &= 0 \\\\\r\n5x - 3y &= 5\r\n\\end{aligned}\r\n$$\r\nNow find the determinants for the system: $4x+5y=4$ and $5x-3y=5$.\r\n$$\r\n\\begin{aligned}\r\nD &= \\begin{vmatrix} 4 & 5 \\\\ 5 & -3 \\end{vmatrix} = (4)(-3) - (5)(5) = -12 - 25 = -37 \\\\\r\nD_x &= \\begin{vmatrix} 4 & 5 \\\\ 5 & -3 \\end{vmatrix} = (4)(-3) - (5)(5) = -12 - 25 = -37 \\\\\r\nD_y &= \\begin{vmatrix} 4 & 4 \\\\ 5 & 5 \\end{vmatrix} = (4)(5) - (4)(5) = 20 - 20 = 0\r\n\\end{aligned}\r\n$$\r\nSolve for x and y:\r\n$$\r\nx = \\frac{D_x}{D} = \\frac{-37}{-37} = 1\r\n$$\r\n$$\r\ny = \\frac{D_y}{D} = \\frac{0}{-37} = 0\r\n$$\r\nThe solution is **$x=1, y=0$**.', 1, '2025-07-25 01:11:38', '2025-09-18 07:21:49');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(142, 13115, '# Grade 10 Optional Mathematics: Angle Between Two Lines\r\n## Unit: Coordinate Geometry | Chapter: Angle Between Two Lines | Complete Solutions\r\n\r\n---\r\n\r\n## 🎯 Chapter Introduction\r\n\r\nWelcome to the comprehensive study guide for finding the **Angle Between Two Straight Lines**. This note is designed to make complex concepts easy to understand, with clear definitions, essential formulas, and detailed, step-by-step solutions to all 29 unique problems from your materials. By the end of this guide, you will have mastered the concepts of slope, parallel and perpendicular lines, homogeneous equations, and finding the equations of lines based on various conditions. 📐\r\n\r\n### Key Learning Areas\r\n1.  **Slope (Gradient):** Calculating the slope of a line from two points or from its equation.\r\n2.  **Angle Formula:** Using the slopes of two lines to find the angle between them.\r\n3.  **Parallel & Perpendicular Lines:** Understanding and applying the conditions for parallelism and perpendicularity.\r\n4.  **Equations of Lines:** Finding the equation of a line that is parallel or perpendicular to another line and passes through a given point or satisfies other conditions.\r\n5.  **Homogeneous Equations:** Finding the angle between the pair of lines represented by a second-degree homogeneous equation.\r\n\r\n---\r\n\r\n## 🔑 Key Concepts & Formulas\r\n\r\nBefore we solve the problems, let\'s understand the main tools we\'ll be using.\r\n\r\n### 1. The Slope of a Straight Line (m)\r\n\r\nThe slope represents the steepness of a line.\r\n\r\n* **From Two Points $(x_1, y_1)$ and $(x_2, y_2)$:**\r\n    $$m = \\frac{y_2 - y_1}{x_2 - x_1}$$\r\n\r\n* **From the General Equation $Ax + By + C = 0$:**\r\n    $$m = -\\frac{\\text{Coefficient of x}}{\\text{Coefficient of y}} = -\\frac{A}{B}$$\r\n\r\n* **From the Slope-Intercept Form $y = mx + c$:**\r\n    The slope is simply `m`.\r\n\r\n---\r\n\r\n### 2. Angle Between Two Lines\r\n\r\n\r\n\r\nIf two lines have slopes $m_1$ and $m_2$, the angle ($\\theta$) between them is given by:\r\n\r\n* **General Formula:**\r\n    $$\\tan\\theta = \\pm \\left( \\frac{m_1 - m_2}{1 + m_1m_2} \\right)$$\r\n    * **Acute Angle:** For the smaller angle (less than 90°), we take the positive value: $\\tan\\theta = \\left| \\frac{m_1 - m_2}{1 + m_1m_2} \\right|$.\r\n    * **Obtuse Angle:** For the larger angle (greater than 90°), we take the negative value.\r\n\r\n---\r\n\r\n### 3. Conditions for Parallel and Perpendicular Lines\r\n\r\n* **Parallel Lines:** Two lines are parallel if and only if their slopes are equal.\r\n    $$m_1 = m_2$$\r\n\r\n* **Perpendicular Lines:** Two lines are perpendicular if and only if the product of their slopes is -1.\r\n    $$m_1 \\cdot m_2 = -1$$\r\n\r\n---\r\n\r\n### 4. Special Equations of Lines\r\n\r\n* **Equation of a line parallel to $Ax + By + C = 0$:**\r\n    The equation will be of the form $Ax + By + k = 0$, where `k` is a new constant found using another condition (like a point on the line).\r\n\r\n* **Equation of a line perpendicular to $Ax + By + C = 0$:**\r\n    The equation will be of the form $Bx - Ay + k = 0$. (Note the swapped coefficients of x and y, and the sign change for the y-coefficient).\r\n\r\n---\r\n\r\n### 5. Pair of Straight Lines (Homogeneous Equation)\r\n\r\nA second-degree homogeneous equation $ax^2 + 2hxy + by^2 = 0$ represents a pair of straight lines passing through the origin.\r\n\r\n* **Angle ($\\theta$) between the pair of lines:**\r\n    $$\\tan\\theta = \\frac{\\pm 2\\sqrt{h^2 - ab}}{a+b}$$\r\n\r\n---\r\n\r\n## 📚 Exercise Solutions\r\n\r\n### Short Questions\r\n\r\n***\r\n\r\n#### Question 1: Finding the Slope of a Perpendicular Line\r\n**Find the slope of the line perpendicular to the straight line 3x - 4y = 10.**\r\n\r\n**Solution:**\r\nFirst, find the slope of the given line $3x - 4y = 10$.\r\n$$m_1 = -\\frac{\\text{Coefficient of x}}{\\text{Coefficient of y}} = -\\frac{3}{-4} = \\frac{3}{4}$$\r\nLet the slope of the perpendicular line be $m_2$. The condition for perpendicularity is $m_1 \\cdot m_2 = -1$.\r\n$$\r\n\\begin{aligned}\r\n\\frac{3}{4} \\cdot m_2 &= -1 \\\\\r\nm_2 &= -\\frac{4}{3}\r\n\\end{aligned}\r\n$$\r\nThe slope of the perpendicular line is **$-\\frac{4}{3}$**.\r\n\r\n***\r\n\r\n#### Question 2: Proving Lines are Parallel\r\n**Prove that the straight line passing through the points (3, -4) and (-2, 6) and the straight line with the equation $2x + y + 3 = 0$ are parallel.**\r\n\r\n**Solution:**\r\n**Step 1: Find the slope of the line from the equation.**\r\nFor the line $2x + y + 3 = 0$:\r\n$$m_1 = -\\frac{\\text{Coefficient of x}}{\\text{Coefficient of y}} = -\\frac{2}{1} = -2$$\r\n**Step 2: Find the slope of the line from the two points.**\r\nFor the points $(3, -4)$ and $(-2, 6)$:\r\n$$m_2 = \\frac{y_2 - y_1}{x_2 - x_1} = \\frac{6 - (-4)}{-2 - 3} = \\frac{10}{-5} = -2$$\r\n**Step 3: Compare the slopes.**\r\nSince $m_1 = m_2 = -2$, the two lines are parallel. **(Proved)**\r\n\r\n***\r\n\r\n#### Question 3: Finding an Unknown for Parallel Lines\r\n**If the lines $3x + my = 5$ and $\\frac{x}{2} + \\frac{y}{3} = 1$ are parallel to each other, find the value of m.**\r\n\r\n**Solution:**\r\n**Step 1: Find the slope of the first line.**\r\nFor $3x + my = 5$, the slope is $m_1 = -\\frac{3}{m}$.\r\n\r\n**Step 2: Find the slope of the second line.**\r\nFirst, convert $\\frac{x}{2} + \\frac{y}{3} = 1$ to the general form by taking the LCM (6):\r\n$$3x + 2y = 6 \\implies 3x + 2y - 6 = 0$$\r\nThe slope is $m_2 = -\\frac{3}{2}$.\r\n\r\n**Step 3: Equate the slopes and solve for m.**\r\nFor parallel lines, $m_1 = m_2$.\r\n$$\r\n\\begin{aligned}\r\n-\\frac{3}{m} &= -\\frac{3}{2} \\\\\r\n\\frac{3}{m} &= \\frac{3}{2} \\\\\r\nm &= 2\r\n\\end{aligned}\r\n$$\r\nThus, **$m = 2$**.\r\n\r\n***\r\n\r\n#### Question 4: Finding an Unknown for Perpendicular Lines\r\n**If the straight line having equation 3x + 4y = 12 and 2x - py = 5 are perpendicular to each other, find the value of p.**\r\n\r\n**Solution:**\r\n**Step 1: Find the slopes of both lines.**\r\nFor $3x + 4y = 12$, the slope is $m_1 = -\\frac{3}{4}$.\r\nFor $2x - py = 5$, the slope is $m_2 = -\\frac{2}{-p} = \\frac{2}{p}$.\r\n\r\n**Step 2: Use the perpendicularity condition and solve for p.**\r\nFor perpendicular lines, $m_1 \\cdot m_2 = -1$.\r\n$$\r\n\\begin{aligned}\r\n\\left(-\\frac{3}{4}\\right) \\cdot \\left(\\frac{2}{p}\\right) &= -1 \\\\\r\n-\\frac{6}{4p} &= -1 \\\\\r\n6 &= 4p \\\\\r\np &= \\frac{6}{4} = \\frac{3}{2}\r\n\\end{aligned}\r\n$$\r\nThus, **$p = \\frac{3}{2}$**.\r\n\r\n***\r\n\r\n#### Question 5: Finding an Unknown for Perpendicular Lines\r\n**If the line passing through points (3, -4) and (-2, k) is perpendicular to the line having equation 5x + y = -3, find the value of k.**\r\n\r\n**Solution:**\r\n**Step 1: Find the slope of the line from the two points.**\r\n$$m_1 = \\frac{k - (-4)}{-2 - 3} = \\frac{k+4}{-5}$$\r\n**Step 2: Find the slope of the line from the equation.**\r\nFor $5x + y = -3$, the slope is $m_2 = -\\frac{5}{1} = -5$.\r\n\r\n**Step 3: Use the perpendicularity condition and solve for k.**\r\n$$\r\n\\begin{aligned}\r\nm_1 \\cdot m_2 &= -1 \\\\\r\n\\left(\\frac{k+4}{-5}\\right) \\cdot (-5) &= -1 \\\\\r\nk+4 &= -1 \\\\\r\nk &= -5\r\n\\end{aligned}\r\n$$\r\nThus, **$k=-5$**.\r\n\r\n***\r\n\r\n#### Question 6: Acute Angle Between Lines\r\n**Find the acute angle between the lines having slope $-\\sqrt{3}$ and $\\sqrt{3}$ respectively.**\r\n\r\n**Solution:**\r\nLet $m_1 = -\\sqrt{3}$ and $m_2 = \\sqrt{3}$. Let $\\theta$ be the acute angle.\r\n$$\r\n\\begin{aligned}\r\n\\tan\\theta &= \\left| \\frac{m_1 - m_2}{1 + m_1m_2} \\right| \\\\\r\n&= \\left| \\frac{-\\sqrt{3} - \\sqrt{3}}{1 + (-\\sqrt{3})(\\sqrt{3})} \\right| \\\\\r\n&= \\left| \\frac{-2\\sqrt{3}}{1 - 3} \\right| \\\\\r\n&= \\left| \\frac{-2\\sqrt{3}}{-2} \\right| \\\\\r\n\\tan\\theta &= \\sqrt{3}\r\n\\end{aligned}\r\n$$\r\nSince $\\tan(60^\\circ) = \\sqrt{3}$, the acute angle is **60°**.\r\n\r\n***\r\n\r\n#### Question 7: Acute Angle Between Lines\r\n**If the slopes of the lines AB and CD are 3 and -2 respectively, find the acute angle between them.**\r\n\r\n**Solution:**\r\nGiven slopes, $m_1 = 3$ and $m_2 = -2$.\r\nLet $\\theta$ be the acute angle.\r\n$$\r\n\\begin{aligned}\r\n\\tan\\theta &= \\left| \\frac{m_1 - m_2}{1 + m_1m_2} \\right| \\\\\r\n&= \\left| \\frac{3 - (-2)}{1 + (3)(-2)} \\right| \\\\\r\n&= \\left| \\frac{5}{1 - 6} \\right| \\\\\r\n&= \\left| \\frac{5}{-5} \\right| = 1\r\n\\end{aligned}\r\n$$\r\nSince $\\tan(45^\\circ) = 1$, the acute angle is **45°**.\r\n\r\n***\r\n\r\n#### Question 8: Finding an Unknown for Parallel Lines\r\n**If the straight lines Px + 3y - 12 = 0 and 4y - 3x + 7 = 0 are parallel to each other, find the value of P.**\r\n\r\n**Solution:**\r\n**Step 1: Find the slopes.**\r\nFor $Px + 3y - 12 = 0$, the slope is $m_1 = -\\frac{P}{3}$.\r\nFor $-3x + 4y + 7 = 0$, the slope is $m_2 = -\\frac{-3}{4} = \\frac{3}{4}$.\r\n\r\n**Step 2: Equate the slopes.**\r\nFor parallel lines, $m_1 = m_2$.\r\n$$\r\n\\begin{aligned}\r\n-\\frac{P}{3} &= \\frac{3}{4} \\\\\r\n-4P &= 9 \\\\\r\nP &= -\\frac{9}{4}\r\n\\end{aligned}\r\n$$\r\nThus, **$P = -9/4$**.\r\n\r\n***\r\n\r\n#### Question 9: Finding an Unknown for Parallel Lines\r\n**If the lines 2x - 8y + 6 = 0 and px - 12y - 4 = 3 are parallel to each other, find the value of p.**\r\n\r\n**Solution:**\r\n**Step 1: Find the slopes.**\r\nFor $2x - 8y + 6 = 0$, the slope is $m_1 = -\\frac{2}{-8} = \\frac{1}{4}$.\r\nFor $px - 12y - 7 = 0$, the slope is $m_2 = -\\frac{p}{-12} = \\frac{p}{12}$.\r\n\r\n**Step 2: Equate the slopes.**\r\n$$\r\n\\begin{aligned}\r\nm_1 &= m_2 \\\\\r\n\\frac{1}{4} &= \\frac{p}{12} \\\\\r\n12 &= 4p \\\\\r\np &= 3\r\n\\end{aligned}\r\n$$\r\nThus, **$p=3$**.\r\n\r\n***\r\n\r\n#### Question 10: Prove Lines are Perpendicular\r\n**Show that the lines $y + \\sqrt{3}x + 4 = 0$ and $x - \\sqrt{3}y = 5$ are perpendicular to each other.**\r\n\r\n**Solution:**\r\n**Step 1: Find the slopes.**\r\nFor $\\sqrt{3}x + y + 4 = 0$, the slope is $m_1 = -\\frac{\\sqrt{3}}{1} = -\\sqrt{3}$.\r\nFor $x - \\sqrt{3}y - 5 = 0$, the slope is $m_2 = -\\frac{1}{-\\sqrt{3}} = \\frac{1}{\\sqrt{3}}$.\r\n\r\n**Step 2: Check the perpendicularity condition.**\r\n$$m_1 \\cdot m_2 = (-\\sqrt{3}) \\cdot \\left(\\frac{1}{\\sqrt{3}}\\right) = -1$$\r\nSince the product of the slopes is -1, the lines are perpendicular. **(Proved)**\r\n\r\n***\r\n\r\n#### Question 11: Prove Lines are Perpendicular\r\n**Show that the lines $ax + by + c = 0$ and $bx - ay + c = 0$ are perpendicular to each other.**\r\n\r\n**Solution:**\r\n**Step 1: Find the slopes.**\r\nFor $ax + by + c = 0$, the slope is $m_1 = -\\frac{a}{b}$.\r\nFor $bx - ay + c = 0$, the slope is $m_2 = -\\frac{b}{-a} = \\frac{b}{a}$.\r\n\r\n**Step 2: Check the perpendicularity condition.**\r\n$$m_1 \\cdot m_2 = \\left(-\\frac{a}{b}\\right) \\cdot \\left(\\frac{b}{a}\\right) = -1$$\r\nSince the product of the slopes is -1, the lines are perpendicular. **(Proved)**\r\n\r\n***\r\n\r\n#### Question 12: Finding an Unknown for Perpendicular Lines\r\n**If the straight lines 2x + 3y + 6 = 0 and ax - 5y + 20 = 0 are perpendicular to each other, find the value of a.**\r\n\r\n**Solution:**\r\n**Step 1: Find the slopes.**\r\nFor $2x + 3y + 6 = 0$, the slope is $m_1 = -\\frac{2}{3}$.\r\nFor $ax - 5y + 20 = 0$, the slope is $m_2 = -\\frac{a}{-5} = \\frac{a}{5}$.\r\n\r\n**Step 2: Use the perpendicularity condition.**\r\n$$\r\n\\begin{aligned}\r\nm_1 \\cdot m_2 &= -1 \\\\\r\n\\left(-\\frac{2}{3}\\right) \\cdot \\left(\\frac{a}{5}\\right) &= -1 \\\\\r\n-\\frac{2a}{15} &= -1 \\\\\r\n2a &= 15 \\\\\r\na &= \\frac{15}{2}\r\n\\end{aligned}\r\n$$\r\nThus, **$a = 15/2$**.\r\n\r\n***\r\n\r\n#### Question 13: Finding an Unknown for Parallel Lines\r\n**If a line passing through the points (4, -p) and (-2, 6) is parallel to the line having equation 2y + 3x = -3, find the value of p.**\r\n\r\n**Solution:**\r\n**Step 1: Find the slopes.**\r\nSlope of the line passing through points (4, -p) and (-2, 6):\r\n$$m_1 = \\frac{6 - (-p)}{-2 - 4} = \\frac{6+p}{-6}$$\r\nSlope of the line $3x + 2y + 3 = 0$:\r\n$$m_2 = -\\frac{3}{2}$$\r\n**Step 2: Equate the slopes.**\r\n$$\r\n\\begin{aligned}\r\n\\frac{6+p}{-6} &= -\\frac{3}{2} \\\\\r\n2(6+p) &= (-3)(-6) \\\\\r\n12 + 2p &= 18 \\\\\r\n2p &= 6 \\\\\r\np &= 3\r\n\\end{aligned}\r\n$$\r\nThus, **$p=3$**.\r\n\r\n***\r\n\r\n### Long Questions\r\n\r\n***\r\n\r\n#### Question 14: Angle between a Pair of Lines\r\n**If an angle between the pair of lines represented by the equation $2x^2 + kxy + 3y^2 = 0$ is 45°, find the positive value of k and also find the separate equations of the lines.**\r\n\r\n**Solution:**\r\n**Step 1: Find k.**\r\nComparing $2x^2 + kxy + 3y^2 = 0$ with $ax^2 + 2hxy + by^2 = 0$, we get:\r\n$a=2$, $b=3$, $2h=k \\implies h=k/2$.\r\nGiven $\\theta = 45^\\circ$.\r\n$$\r\n\\begin{aligned}\r\n\\tan\\theta &= \\frac{\\pm 2\\sqrt{h^2 - ab}}{a+b} \\\\\r\n\\tan45^\\circ &= \\frac{\\pm 2\\sqrt{(k/2)^2 - (2)(3)}}{2+3} \\\\\r\n1 &= \\frac{\\pm 2\\sqrt{k^2/4 - 6}}{5} \\\\\r\n5 &= \\pm \\sqrt{k^2 - 24} \\quad (\\text{since } \\sqrt{4}=2)\r\n\\end{aligned}\r\n$$Squaring both sides:$$\r\n\\begin{aligned}\r\n25 &= k^2 - 24 \\\\\r\nk^2 &= 49 \\\\\r\nk &= \\pm 7\r\n\\end{aligned}\r\n$$\r\nThe positive value is **$k=7$**.\r\n\r\n**Step 2: Find the separate equations.**\r\nSubstitute $k=7$ into the original equation:\r\n$$\r\n\\begin{aligned}\r\n2x^2 + 7xy + 3y^2 &= 0 \\\\\r\n2x^2 + 6xy + xy + 3y^2 &= 0 \\\\\r\n2x(x+3y) + y(x+3y) &= 0 \\\\\r\n(2x+y)(x+3y) &= 0\r\n\\end{aligned}\r\n$$\r\nThe separate equations are **$2x+y=0$** and **$x+3y=0$**.\r\n\r\n***\r\n\r\n#### Question 15: Equation of a Line through Intersection\r\n**Find the equation of a straight line passing through the point of intersection of the lines $3x + 4y = 7$ and $5x - 2y = 3$ and perpendicular to the line $2x + 3y = 5$.**\r\n\r\n**Solution:**\r\n**Step 1: Find the point of intersection.**\r\nFrom $3x + 4y = 7$, we get $x = \\frac{7-4y}{3}$.\r\nSubstitute into the second equation:\r\n$$\r\n\\begin{aligned}\r\n5\\left(\\frac{7-4y}{3}\\right) - 2y &= 3 \\\\\r\n35 - 20y - 6y &= 9 \\\\\r\n35 - 26y &= 9 \\\\\r\n26 &= 26y \\\\\r\ny &= 1\r\n\\end{aligned}\r\n$$\r\nNow find x: $x = \\frac{7-4(1)}{3} = \\frac{3}{3} = 1$. The intersection point is (1, 1).\r\n\r\n**Step 2: Find the equation of the required line.**\r\nThe slope of $2x + 3y = 5$ is $m_1 = -2/3$.\r\nThe required line is perpendicular, so its slope is $m_2 = -1/m_1 = 3/2$.\r\nThe line passes through (1, 1) with slope 3/2. Using point-slope form:\r\n$$\r\n\\begin{aligned}\r\ny - 1 &= \\frac{3}{2}(x - 1) \\\\\r\n2y - 2 &= 3x - 3 \\\\\r\n3x - 2y - 1 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$3x - 2y - 1 = 0$**.\r\n\r\n***\r\n\r\n#### Question 16: Equation of a Perpendicular Bisector\r\n**Find the equation of the perpendicular bisector of the line segment joining the points A(3, -7) and B(-5, 3).**\r\n\r\n**Solution:**\r\n**Step 1: Find the midpoint of AB.**\r\n$$Midpoint = \\left( \\frac{3 + (-5)}{2}, \\frac{-7+3}{2} \\right) = (-1, -2)$$\r\n**Step 2: Find the slope of line AB.**\r\n$$m_{AB} = \\frac{3 - (-7)}{-5 - 3} = \\frac{10}{-8} = -\\frac{5}{4}$$\r\n**Step 3: Find the slope of the perpendicular bisector.**\r\n$$m_{\\perp} = -\\frac{1}{m_{AB}} = -\\frac{1}{-5/4} = \\frac{4}{5}$$\r\n**Step 4: Find the equation of the bisector.**\r\nThe line passes through $(-1, -2)$ with slope $4/5$.\r\n$$\r\n\\begin{aligned}\r\ny - (-2) &= \\frac{4}{5}(x - (-1)) \\\\\r\n5(y+2) &= 4(x+1) \\\\\r\n5y + 10 &= 4x + 4 \\\\\r\n4x - 5y - 6 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$4x - 5y - 6 = 0$**.\r\n\r\n***\r\n\r\n#### Question 17: Equation of a Line through a Centroid\r\n**A(3, 2), B(1, -1) and C(5, -5) are the vertices of a triangle ABC. Find the equation of a straight line passing through the centroid of triangle ABC and parallel to the side BC.**\r\n\r\n**Solution:**\r\n**Step 1: Find the centroid of triangle ABC.**\r\n$$Centroid = \\left( \\frac{3+1+5}{3}, \\frac{2+(-1)+(-5)}{3} \\right) = \\left( \\frac{9}{3}, \\frac{-4}{3} \\right) = (3, -4/3)$$\r\n**Step 2: Find the slope of the side BC.**\r\n$$m_{BC} = \\frac{-5 - (-1)}{5 - 1} = \\frac{-4}{4} = -1$$\r\n**Step 3: Find the equation of the required line.**\r\nThe required line is parallel to BC, so its slope is also $m = -1$.\r\nIt passes through the centroid $(3, -4/3)$.\r\n$$\r\n\\begin{aligned}\r\ny - (-\\frac{4}{3}) &= -1(x - 3) \\\\\r\ny + \\frac{4}{3} &= -x + 3 \\\\\r\n3y + 4 &= -3x + 9 \\\\\r\n3x + 3y - 5 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$3x + 3y - 5 = 0$**.\r\n\r\n***\r\n\r\n#### Question 18: Equation of a Line passing through Centroid\r\n**Find the equation of the line passing through the centroid of triangle PQR with vertices P(3, 3), Q(-2, -6) and R(5, -3) and parallel to the line QR.**\r\n\r\n**Solution:**\r\n**Step 1: Find the centroid of triangle PQR.**\r\n$$Centroid = \\left( \\frac{3+(-2)+5}{3}, \\frac{3+(-6)+(-3)}{3} \\right) = \\left( \\frac{6}{3}, \\frac{-6}{3} \\right) = (2, -2)$$\r\n**Step 2: Find the slope of the side QR.**\r\n$$m_{QR} = \\frac{-3 - (-6)}{5 - (-2)} = \\frac{3}{7}$$\r\n**Step 3: Find the equation of the required line.**\r\nThe required line is parallel to QR, so its slope is also $m = 3/7$.\r\nIt passes through the centroid $(2, -2)$.\r\n$$\r\n\\begin{aligned}\r\ny - (-2) &= \\frac{3}{7}(x - 2) \\\\\r\n7(y+2) &= 3(x-2) \\\\\r\n7y + 14 &= 3x - 6 \\\\\r\n3x - 7y - 20 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$3x - 7y - 20 = 0$**.', 1, '2025-07-25 01:11:38', '2025-09-18 07:21:49'),
(143, 13117, '# Grade 10 Optional Mathematics: The Circle\r\n## Unit: Coordinate Geometry | Chapter: Circle | Complete Solutions\r\n\r\n---\r\n\r\n## 🎯 Chapter Introduction\r\n\r\nThis guide offers a complete overview of the **Circle** chapter in Coordinate Geometry. It is designed to make the concepts easy to understand, with clear definitions, essential formulas, and detailed, step-by-step solutions to a wide range of problems. By the end of this note, you will have mastered the different forms of the circle equation and how to find a circle\'s center, radius, and equation based on various given conditions. ⚪\r\n\r\n### Key Learning Areas\r\n1.  **Standard Equation of a Circle:** Understanding and using the center-radius form.\r\n2.  **General Equation of a Circle:** Identifying the center and radius from the general form.\r\n3.  **Diameter Form:** Finding the equation of a circle given the endpoints of a diameter.\r\n4.  **Problem Solving:** Finding the equation of a circle based on various conditions, such as passing through points, having a specific center, or being concentric with another circle.\r\n\r\n---\r\n\r\n## 🔑 Key Concepts & Formulas\r\n\r\nBefore we solve the problems, let\'s understand the fundamental equations of a circle.\r\n\r\n### 1. Standard Equation of a Circle (Center-Radius Form)\r\n\r\nThis is the most common form of a circle\'s equation. It directly uses the circle\'s center and radius.\r\n\r\n* **Formula:**\r\n    $$(x - h)^2 + (y - k)^2 = r^2$$\r\n    Where:\r\n    * `(h, k)` are the coordinates of the center of the circle.\r\n    * `r` is the radius of the circle.\r\n\r\n\r\n\r\n---\r\n\r\n### 2. General Equation of a Circle\r\n\r\nThis form is derived by expanding the standard equation. It is useful for identifying a circle\'s properties when the equation is not in center-radius form.\r\n\r\n* **Formula:**\r\n    $$x^2 + y^2 + 2gx + 2fy + c = 0$$\r\n\r\n* **Finding the Center and Radius from the General Form:**\r\n    * **Center:** $(-g, -f)$\r\n    * **Radius:** $r = \\sqrt{g^2 + f^2 - c}$\r\n\r\n    To find `g`, `f`, and `c`, you compare the given equation to the general form. For example, if the equation is $x^2 + y^2 - 10x - 4y - 7 = 0$:\r\n    * `2gx = -10x`  => `g = -5`\r\n    * `2fy = -4y`  => `f = -2`\r\n    * `c = -7`\r\n\r\n---\r\n\r\n### 3. Equation of a Circle in Diameter Form\r\n\r\nIf you know the coordinates of the two endpoints of a circle\'s diameter, you can use this special formula.\r\n\r\n* **Formula:**\r\n    $$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$\r\n    Where $(x_1, y_1)$ and $(x_2, y_2)$ are the coordinates of the endpoints of a diameter.\r\n\r\n\r\n\r\n---\r\n\r\n### 4. Midpoint Formula\r\n\r\nThis is useful for finding the center of a circle if you know the endpoints of a diameter.\r\n\r\n* **Formula:**\r\n    $$Midpoint = \\left( \\frac{x_1 + x_2}{2}, \\frac{y_1 + y_2}{2} \\right)$$\r\n\r\n---\r\n\r\n## 📚 Exercise Solutions\r\n\r\n### Short Questions\r\n\r\n***\r\n\r\n#### Question 1: Equation from Diameter Endpoints\r\n**The endpoints of a diameter of a circle are (0, 6) and (8, 0). Find the equation of the circle.**\r\n\r\n**Solution:**\r\nLet $(x_1, y_1) = (0, 6)$ and $(x_2, y_2) = (8, 0)$.\r\nUsing the diameter form of the circle equation:\r\n$$\r\n\\begin{aligned}\r\n(x - x_1)(x - x_2) + (y - y_1)(y - y_2) &= 0 \\\\\r\n(x - 0)(x - 8) + (y - 6)(y - 0) &= 0 \\\\\r\nx(x - 8) + y(y - 6) &= 0 \\\\\r\nx^2 - 8x + y^2 - 6y &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 8x - 6y = 0$**.\r\n\r\n***\r\n\r\n#### Question 2: Equation from Diameter Endpoints\r\n**The endpoints of a diameter of a circle are (3, 0) and (0, 4). Find the equation of the circle.**\r\n\r\n**Solution:**\r\nLet $(x_1, y_1) = (3, 0)$ and $(x_2, y_2) = (0, 4)$.\r\nUsing the diameter form:\r\n$$\r\n\\begin{aligned}\r\n(x - 3)(x - 0) + (y - 0)(y - 4) &= 0 \\\\\r\nx(x - 3) + y(y - 4) &= 0 \\\\\r\nx^2 - 3x + y^2 - 4y &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 3x - 4y = 0$**.\r\n\r\n***\r\n\r\n#### Question 3: Equation from Diameter Endpoints\r\n**Find the equation of the circle having (1, 3) and (2, -5) as the endpoints of a diameter.**\r\n\r\n**Solution:**\r\nLet $(x_1, y_1) = (1, 3)$ and $(x_2, y_2) = (2, -5)$.\r\nUsing the diameter form:\r\n$$\r\n\\begin{aligned}\r\n(x - 1)(x - 2) + (y - 3)(y - (-5)) &= 0 \\\\\r\n(x^2 - 2x - x + 2) + (y - 3)(y + 5) &= 0 \\\\\r\nx^2 - 3x + 2 + y^2 + 5y - 3y - 15 &= 0 \\\\\r\nx^2 + y^2 - 3x + 2y - 13 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 3x + 2y - 13 = 0$**.\r\n\r\n***\r\n\r\n#### Question 4: Equation from Diameter Endpoints\r\n**Find the equation of the circle having the endpoints of a diameter A(5, 6) and B(3, 4).**\r\n\r\n**Solution:**\r\nLet $(x_1, y_1) = (5, 6)$ and $(x_2, y_2) = (3, 4)$.\r\nUsing the diameter form:\r\n$$\r\n\\begin{aligned}\r\n(x - 5)(x - 3) + (y - 6)(y - 4) &= 0 \\\\\r\n(x^2 - 3x - 5x + 15) + (y^2 - 4y - 6y + 24) &= 0 \\\\\r\nx^2 - 8x + 15 + y^2 - 10y + 24 &= 0 \\\\\r\nx^2 + y^2 - 8x - 10y + 39 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 8x - 10y + 39 = 0$**.\r\n\r\n***\r\n\r\n#### Question 5: Equation from Diameter Endpoints\r\n**Find the equation of the circle whose endpoints of a diameter are A(3, 4) and B(2, -7).**\r\n\r\n**Solution:**\r\nLet $(x_1, y_1) = (3, 4)$ and $(x_2, y_2) = (2, -7)$.\r\nUsing the diameter form:\r\n$$\r\n\\begin{aligned}\r\n(x - 3)(x - 2) + (y - 4)(y - (-7)) &= 0 \\\\\r\n(x^2 - 2x - 3x + 6) + (y - 4)(y + 7) &= 0 \\\\\r\nx^2 - 5x + 6 + y^2 + 7y - 4y - 28 &= 0 \\\\\r\nx^2 + y^2 - 5x + 3y - 22 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 5x + 3y - 22 = 0$**.\r\n\r\n***\r\n\r\n#### Question 6: Center from Intersecting Diameters\r\n**Find the coordinates of the center of a circle having equations of two diameters $x + y = 5$ and $2x - y = 1$.**\r\n\r\n**Solution:**\r\nThe center of a circle is the point of intersection of its diameters. We solve the system of linear equations.\r\n$$\r\n\\begin{aligned}\r\nx + y &= 5 \\quad &(1) \\\\\r\n2x - y &= 1 \\quad &(2)\r\n\\end{aligned}\r\n$$\r\nAdding equation (1) and (2):\r\n$$\r\n\\begin{aligned}\r\n(x+y) + (2x-y) &= 5 + 1 \\\\\r\n3x &= 6 \\\\\r\nx &= 2\r\n\\end{aligned}\r\n$$\r\nSubstitute $x=2$ into equation (1):\r\n$$\r\n2 + y = 5 \\implies y = 3\r\n$$\r\nThe center of the circle is **(2, 3)**.\r\n\r\n***\r\n\r\n#### Question 7: Center and Radius from General Equation\r\n**Find the center and radius of a circle having the equation $x^2 + y^2 - 10x - 4y - 7 = 0$.**\r\n\r\n**Solution:**\r\nCompare the given equation to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = -10x &\\implies g = -5 \\\\\r\n2fy = -4y &\\implies f = -2 \\\\\r\nc &= -7\r\n\\end{aligned}\r\n$$\r\n**Center:** $(-g, -f) = (-(-5), -(-2)) = (5, 2)$.\r\n**Radius:** $r = \\sqrt{g^2 + f^2 - c} = \\sqrt{(-5)^2 + (-2)^2 - (-7)} = \\sqrt{25 + 4 + 7} = \\sqrt{36} = 6$.\r\n\r\nThe center is **(5, 2)** and the radius is **6 units**.\r\n\r\n***\r\n\r\n#### Question 8: Finding an Unknown from Radius\r\n**If the radius of the circle $x^2 + y^2 - 4x - 6y - k = 0$ is 4 units, find the value of k.**\r\n\r\n**Solution:**\r\nCompare the equation to $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = -4x &\\implies g = -2 \\\\\r\n2fy = -6y &\\implies f = -3 \\\\\r\nc &= -k\r\n\\end{aligned}\r\n$$\r\nUsing the radius formula, with $r=4$:\r\n$$\r\n\\begin{aligned}\r\nr &= \\sqrt{g^2 + f^2 - c} \\\\\r\n4 &= \\sqrt{(-2)^2 + (-3)^2 - (-k)} \\\\\r\n4 &= \\sqrt{4 + 9 + k} \\\\\r\n4 &= \\sqrt{13 + k}\r\n\\end{aligned}\r\n$$\r\nSquaring both sides:\r\n$$\r\n16 = 13 + k \\implies k = 3\r\n$$\r\nThus, the value of **$k = 3$**.\r\n\r\n***\r\n\r\n#### Question 9: Finding Unknowns from Center\r\n**If the center of the circle $x^2 + y^2 - ax - by - 12 = 0$ is (2, 3), find the value of a and b.**\r\n\r\n**Solution:**\r\nCompare the equation to $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = -ax &\\implies g = -a/2 \\\\\r\n2fy = -by &\\implies f = -b/2\r\n\\end{aligned}\r\n$$\r\nThe center is given by $(-g, -f)$.\r\n$$\r\nCenter = (-(-a/2), -(-b/2)) = (a/2, b/2)\r\n$$\r\nWe are given that the center is (2, 3).\r\n$$\r\n\\begin{aligned}\r\na/2 &= 2 \\implies a = 4 \\\\\r\nb/2 &= 3 \\implies b = 6\r\n\\end{aligned}\r\n$$\r\nThus, **$a=4$ and $b=6$**.\r\n\r\n***\r\n\r\n#### Question 10: Finding an Unknown from Center\r\n**Find the value of m when the center of a circle with equation $x^2 + y^2 - mx + 6y - 12 = 0$ is (-2, -3).**\r\n\r\n**Solution:**\r\nCompare the equation to $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = -mx &\\implies g = -m/2 \\\\\r\n2fy = 6y &\\implies f = 3\r\n\\end{aligned}\r\n$$\r\nThe center is $(-g, -f) = (-(-m/2), -3) = (m/2, -3)$.\r\nWe are given that the center is $(-2, -3)$.\r\nEquating the x-coordinates:\r\n$$\r\n\\frac{m}{2} = -2 \\implies m = -4\r\n$$\r\nThus, **$m = -4$**.\r\n\r\n***\r\n\r\n#### Question 11: Equation from Center and Radius\r\n**Find the equation of a circle having centre (1, -2) and diameter $2\\sqrt{5}$ units.**\r\n\r\n**Solution:**\r\nCenter $(h, k) = (1, -2)$.\r\nDiameter is $2\\sqrt{5}$, so the radius is $r = \\sqrt{5}$.\r\nUsing the standard form $(x-h)^2 + (y-k)^2 = r^2$:\r\n$$\r\n\\begin{aligned}\r\n(x - 1)^2 + (y - (-2))^2 &= (\\sqrt{5})^2 \\\\\r\n(x^2 - 2x + 1) + (y^2 + 4y + 4) &= 5 \\\\\r\nx^2 + y^2 - 2x + 4y + 5 &= 5\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 2x + 4y = 0$**.\r\n\r\n***\r\n\r\n#### Question 12: Equation from Center and Radius\r\n**Find the equation of a circle having centre (2, 3) and radius 5 units.**\r\n\r\n**Solution:**\r\nCenter $(h, k) = (2, 3)$ and radius $r = 5$.\r\nUsing the standard form:\r\n$$\r\n\\begin{aligned}\r\n(x - 2)^2 + (y - 3)^2 &= 5^2 \\\\\r\n(x^2 - 4x + 4) + (y^2 - 6y + 9) &= 25 \\\\\r\nx^2 + y^2 - 4x - 6y + 13 &= 25\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 4x - 6y - 12 = 0$**.\r\n\r\n***\r\n\r\n#### Question 13: Center from General Equation\r\n**Find the coordinates of the center of a circle having an equation $x^2 + y^2 + 4x - 5 = 0$.**\r\n\r\n**Solution:**\r\nCompare with $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = 4x &\\implies g = 2 \\\\\r\n2fy = 0y &\\implies f = 0\r\n\\end{aligned}\r\n$$\r\nThe center is $(-g, -f) = (-2, 0)$.\r\nThe center is **(-2, 0)**.\r\n\r\n***\r\n\r\n#### Question 14: Equation from Center and Radius\r\n**Find the equation of a circle having the centre (-1, -3) and radius 3 units.**\r\n\r\n**Solution:**\r\nCenter $(h, k) = (-1, -3)$ and radius $r=3$.\r\nUsing the standard form:\r\n$$\r\n\\begin{aligned}\r\n(x - (-1))^2 + (y - (-3))^2 &= 3^2 \\\\\r\n(x+1)^2 + (y+3)^2 &= 9 \\\\\r\n(x^2 + 2x + 1) + (y^2 + 6y + 9) &= 9 \\\\\r\nx^2 + y^2 + 2x + 6y + 1 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 + 2x + 6y + 1 = 0$**.\r\n\r\n***\r\n\r\n#### Question 15: Center from General Equation\r\n**Find the center of the circle whose equation is $x^2 + y^2 - 6x + 2y + 1 = 0$.**\r\n\r\n**Solution:**\r\nCompare with $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = -6x &\\implies g = -3 \\\\\r\n2fy = 2y &\\implies f = 1\r\n\\end{aligned}\r\n$$\r\nThe center is $(-g, -f) = (-(-3), -1) = (3, -1)$.\r\nThe center is **(3, -1)**.\r\n\r\n---\r\n\r\n### Long Questions\r\n\r\n***\r\n\r\n#### Question 16: Equation of Circle Passing Through Center of Another\r\n**Circle A with centre X(-4, 5) passes through the centre Y of circle B. If the equation of circle B is $x^2 + y^2 - 4x + 6y - 12 = 0$, find the equation of circle A.**\r\n\r\n**Solution:**\r\n**Step 1: Find the center of circle B.**\r\nCompare $x^2 + y^2 - 4x + 6y - 12 = 0$ with the general form.\r\n$g = -2$, $f=3$.\r\nCenter of B is $Y = (-g, -f) = (2, -3)$.\r\n\r\n**Step 2: Find the radius of circle A.**\r\nThe radius of circle A is the distance between its center X(-4, 5) and the point it passes through, Y(2, -3).\r\n$$\r\n\\begin{aligned}\r\nr^2 &= (x_2 - x_1)^2 + (y_2 - y_1)^2 \\\\\r\n&= (2 - (-4))^2 + (-3 - 5)^2 \\\\\r\n&= (6)^2 + (-8)^2 \\\\\r\n&= 36 + 64 = 100\r\n\\end{aligned}\r\n$$\r\n**Step 3: Find the equation of circle A.**\r\nThe center of circle A is X(h, k) = (-4, 5) and $r^2=100$.\r\n$$\r\n\\begin{aligned}\r\n(x - (-4))^2 + (y - 5)^2 &= 100 \\\\\r\n(x+4)^2 + (y-5)^2 &= 100 \\\\\r\nx^2 + 8x + 16 + y^2 - 10y + 25 &= 100 \\\\\r\nx^2 + y^2 + 8x - 10y - 59 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 + 8x - 10y - 59 = 0$**.\r\n\r\n***\r\n\r\n#### Question 17: Equation of Circle Passing Through Center of Another\r\n**Find the equation of a circle with centre (-1, 2) and passing through the centre of the circle $x^2 + y^2 - 6x - 10y - 2 = 0$.**\r\n\r\n**Solution:**\r\n**Step 1: Find the center of the given circle.**\r\nComparing $x^2 + y^2 - 6x - 10y - 2 = 0$ to the general form.\r\n$g=-3$, $f=-5$.\r\nThe center is $(-g, -f) = (3, 5)$.\r\n\r\n**Step 2: Find the radius of the required circle.**\r\nThe required circle has center $(h,k)=(-1,2)$ and passes through $(3,5)$. The radius is the distance between these two points.\r\n$$\r\n\\begin{aligned}\r\nr^2 &= (3 - (-1))^2 + (5 - 2)^2 \\\\\r\n&= (4)^2 + (3)^2 \\\\\r\n&= 16 + 9 = 25\r\n\\end{aligned}\r\n$$\r\n**Step 3: Find the equation of the required circle.**\r\nCenter is $(-1, 2)$ and $r^2=25$.\r\n$$\r\n\\begin{aligned}\r\n(x - (-1))^2 + (y - 2)^2 &= 25 \\\\\r\n(x+1)^2 + (y-2)^2 &= 25 \\\\\r\nx^2 + 2x + 1 + y^2 - 4y + 4 &= 25 \\\\\r\nx^2 + y^2 + 2x - 4y - 20 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 + 2x - 4y - 20 = 0$**.\r\n\r\n***\r\n\r\n#### Question 18: Finding the Other End of a Diameter\r\n**If one endpoint of a diameter of the circle $x^2 + y^2 + 4x - 6y + 8 = 0$ is (-4, 4), find the coordinates of the other endpoint.**\r\n\r\n**Solution:**\r\n**Step 1: Find the center of the circle.**\r\nComparing the equation to the general form: $g=2$, $f=-3$.\r\nThe center is $(-g, -f) = (-2, 3)$.\r\n\r\n**Step 2: Use the midpoint formula.**\r\nThe center of a circle is the midpoint of its diameter. Let the other endpoint be $(x_2, y_2)$.\r\n$$\r\n\\begin{aligned}\r\n\\text{Center} &= \\left( \\frac{x_1+x_2}{2}, \\frac{y_1+y_2}{2} \\right) \\\\\r\n(-2, 3) &= \\left( \\frac{-4+x_2}{2}, \\frac{4+y_2}{2} \\right)\r\n\\end{aligned}\r\n$$\r\nEquating coordinates:\r\n$$\r\n-2 = \\frac{-4+x_2}{2} \\implies -4 = -4+x_2 \\implies x_2=0\r\n$$\r\n$$\r\n3 = \\frac{4+y_2}{2} \\implies 6 = 4+y_2 \\implies y_2=2\r\n$$\r\nThe coordinates of the other end are **(0, 2)**.\r\n\r\n***\r\n\r\n#### Question 19: Equation of a Concentric Circle\r\n**Find the equation of a circle concentric with the circle $2x^2 + 2y^2 + 4x - 2y + 1 = 0$ and passing through the point (4,-2).**\r\n\r\n**Solution:**\r\n**Step 1: Find the center of the given circle.**\r\nFirst, divide the equation by 2 to get the standard general form:\r\n$x^2 + y^2 + 2x - y + \\frac{1}{2} = 0$.\r\nComparing with the general form, $g=1$ and $f=-1/2$.\r\nThe center is $(-g, -f) = (-1, 1/2)$.\r\n\r\n**Step 2: Find the radius of the required circle.**\r\nSince the required circle is concentric, its center is also $(-1, 1/2)$. It passes through $(4, -2)$. The radius is the distance between these points.\r\n$$\r\n\\begin{aligned}\r\nr^2 &= (4 - (-1))^2 + (-2 - \\frac{1}{2})^2 \\\\\r\n&= (5)^2 + (-\\frac{5}{2})^2 \\\\\r\n&= 25 + \\frac{25}{4} = \\frac{100+25}{4} = \\frac{125}{4}\r\n\\end{aligned}\r\n$$\r\n**Step 3: Write the equation of the required circle.**\r\nCenter is $(-1, 1/2)$ and $r^2 = 125/4$.\r\n$$\r\n\\begin{aligned}\r\n(x - (-1))^2 + (y - \\frac{1}{2})^2 &= \\frac{125}{4} \\\\\r\n(x+1)^2 + (y - \\frac{1}{2})^2 &= \\frac{125}{4} \\\\\r\nx^2+2x+1 + y^2 - y + \\frac{1}{4} &= \\frac{125}{4}\r\n\\end{aligned}\r\n$$\r\nMultiply by 4 to clear the fraction:\r\n$$\r\n\\begin{aligned}\r\n4x^2 + 8x + 4 + 4y^2 - 4y + 1 &= 125 \\\\\r\n4x^2 + 4y^2 + 8x - 4y - 120 &= 0 \\\\\r\nx^2 + y^2 + 2x - y - 30 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 + 2x - y - 30 = 0$**.\r\n\r\n***\r\n\r\n#### Question 20: Equation of a Circle from Two Points and Center on a Line\r\n**Find the equation of the circle which passes through the points (4, 1) and (6, 5) and has its centre on the line 4x + y = 16.**\r\n\r\n**Solution:**\r\nLet the center be $(h,k)$. Since it lies on the line $4x+y=16$, we have $4h+k=16 \\implies k = 16-4h$.\r\nThe distance from the center to any point on the circle is the radius. So, the distance from $(h,k)$ to $(4,1)$ is the same as to $(6,5)$.\r\n$$\r\n\\begin{aligned}\r\n(h-4)^2 + (k-1)^2 &= (h-6)^2 + (k-5)^2 \\quad (\\text{since } r^2=r^2) \\\\\r\nh^2 - 8h + 16 + k^2 - 2k + 1 &= h^2 - 12h + 36 + k^2 - 10k + 25 \\\\\r\n-8h - 2k + 17 &= -12h - 10k + 61 \\\\\r\n4h + 8k &= 44 \\\\\r\nh + 2k &= 11\r\n\\end{aligned}\r\n$$\r\nNow substitute $k = 16-4h$ into this equation:\r\n$$\r\n\\begin{aligned}\r\nh + 2(16-4h) &= 11 \\\\\r\nh + 32 - 8h &= 11 \\\\\r\n-7h &= -21 \\\\\r\nh &= 3\r\n\\end{aligned}\r\n$$\r\nNow find k: $k = 16 - 4(3) = 16 - 12 = 4$.\r\nThe center is $(3,4)$.\r\nFind the radius squared, $r^2$, using point $(4,1)$:\r\n$$\r\nr^2 = (4-3)^2 + (1-4)^2 = 1^2 + (-3)^2 = 1+9=10\r\n$$\r\nThe equation is $(x-3)^2 + (y-4)^2 = 10$, which expands to **$x^2 + y^2 - 6x - 8y + 15 = 0$**.\r\n\r\n***\r\n\r\n#### Question 21: Equation of a Circle from Three Points\r\n**Find the equation of the circle which passes through the points P(5, 7), Q(6, 6) and R(2, -2).**\r\n\r\n**Solution:**\r\nLet the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$.\r\nThe squared distance from the center $(h,k)$ to each point is equal.\r\n**From P and Q:**\r\n$$\r\n\\begin{aligned}\r\n(h-5)^2 + (k-7)^2 &= (h-6)^2 + (k-6)^2 \\\\\r\nh^2-10h+25 + k^2-14k+49 &= h^2-12h+36 + k^2-12k+36 \\\\\r\n-10h-14k+74 &= -12h-12k+72 \\\\\r\n2h - 2k + 2 &= 0 \\implies h - k + 1 = 0 \\implies h = k-1 \\quad \\text{(Eq. 1)}\r\n\\end{aligned}\r\n$$\r\n**From Q and R:**\r\n$$\r\n\\begin{aligned}\r\n(h-6)^2 + (k-6)^2 &= (h-2)^2 + (k-(-2))^2 \\\\\r\nh^2-12h+36 + k^2-12k+36 &= h^2-4h+4 + k^2+4k+4 \\\\\r\n-12h-12k+72 &= -4h+4k+8 \\\\\r\n64 &= 8h + 16k \\\\\r\n8 &= h + 2k \\quad \\text{(Eq. 2)}\r\n\\end{aligned}\r\n$$\r\n**Solve for h and k:**\r\nSubstitute Eq. 1 into Eq. 2:\r\n$8 = (k-1) + 2k \\implies 9 = 3k \\implies k=3$.\r\nThen, $h = 3-1=2$. The center is $(h,k)=(2,3)$.\r\n\r\n**Find $r^2$ using point P(5,7):**\r\n$r^2 = (5-2)^2 + (7-3)^2 = 3^2 + 4^2 = 9+16=25$.\r\n\r\n**Write the equation:**\r\n$$(x-2)^2 + (y-3)^2 = 25$$\r\nThe required equation is **$x^2 + y^2 - 4x - 6y - 12 = 0$**.', 1, '2025-07-25 01:11:38', '2025-09-18 07:21:49');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(144, 13116, '# Grade 10 Optional Mathematics: The Circle\r\n## Unit: Coordinate Geometry | Chapter: Circle | Complete Solutions\r\n\r\n---\r\n\r\n## 🎯 Chapter Introduction\r\n\r\nThis guide offers a complete overview of the **Circle** chapter in Coordinate Geometry. It is designed to make the concepts easy to understand, with clear definitions, essential formulas, and detailed, step-by-step solutions to a wide range of problems. By the end of this note, you will have mastered the different forms of the circle equation and how to find a circle\'s center, radius, and equation based on various given conditions. ⚪\r\n\r\n### Key Learning Areas\r\n1.  **Standard Equation of a Circle:** Understanding and using the center-radius form.\r\n2.  **General Equation of a Circle:** Identifying the center and radius from the general form.\r\n3.  **Diameter Form:** Finding the equation of a circle given the endpoints of a diameter.\r\n4.  **Problem Solving:** Finding the equation of a circle based on various conditions, such as passing through points, having a specific center, or being concentric with another circle.\r\n\r\n---\r\n\r\n## 🔑 Key Concepts & Formulas\r\n\r\nBefore we solve the problems, let\'s understand the fundamental equations of a circle.\r\n\r\n### 1. Standard Equation of a Circle (Center-Radius Form)\r\n\r\nThis is the most common form of a circle\'s equation. It directly uses the circle\'s center and radius.\r\n\r\n* **Formula:**\r\n    $$(x - h)^2 + (y - k)^2 = r^2$$\r\n    Where:\r\n    * `(h, k)` are the coordinates of the center of the circle.\r\n    * `r` is the radius of the circle.\r\n\r\n\r\n\r\n---\r\n\r\n### 2. General Equation of a Circle\r\n\r\nThis form is derived by expanding the standard equation. It is useful for identifying a circle\'s properties when the equation is not in center-radius form.\r\n\r\n* **Formula:**\r\n    $$x^2 + y^2 + 2gx + 2fy + c = 0$$\r\n\r\n* **Finding the Center and Radius from the General Form:**\r\n    * **Center:** $(-g, -f)$\r\n    * **Radius:** $r = \\sqrt{g^2 + f^2 - c}$\r\n\r\n    To find `g`, `f`, and `c`, you compare the given equation to the general form. For example, if the equation is $x^2 + y^2 - 10x - 4y - 7 = 0$:\r\n    * `2gx = -10x`  => `g = -5`\r\n    * `2fy = -4y`  => `f = -2`\r\n    * `c = -7`\r\n\r\n---\r\n\r\n### 3. Equation of a Circle in Diameter Form\r\n\r\nIf you know the coordinates of the two endpoints of a circle\'s diameter, you can use this special formula.\r\n\r\n* **Formula:**\r\n    $$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$\r\n    Where $(x_1, y_1)$ and $(x_2, y_2)$ are the coordinates of the endpoints of a diameter.\r\n\r\n\r\n\r\n---\r\n\r\n### 4. Midpoint Formula\r\n\r\nThis is useful for finding the center of a circle if you know the endpoints of a diameter.\r\n\r\n* **Formula:**\r\n    $$Midpoint = \\left( \\frac{x_1 + x_2}{2}, \\frac{y_1 + y_2}{2} \\right)$$\r\n\r\n---\r\n\r\n## 📚 Exercise Solutions\r\n\r\n### Short Questions\r\n\r\n***\r\n\r\n#### Question 1: Equation from Diameter Endpoints\r\n**The endpoints of a diameter of a circle are (0, 6) and (8, 0). Find the equation of the circle.**\r\n\r\n**Solution:**\r\nLet $(x_1, y_1) = (0, 6)$ and $(x_2, y_2) = (8, 0)$.\r\nUsing the diameter form of the circle equation:\r\n$$\r\n\\begin{aligned}\r\n(x - x_1)(x - x_2) + (y - y_1)(y - y_2) &= 0 \\\\\r\n(x - 0)(x - 8) + (y - 6)(y - 0) &= 0 \\\\\r\nx(x - 8) + y(y - 6) &= 0 \\\\\r\nx^2 - 8x + y^2 - 6y &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 8x - 6y = 0$**.\r\n\r\n***\r\n\r\n#### Question 2: Equation from Diameter Endpoints\r\n**The endpoints of a diameter of a circle are (3, 0) and (0, 4). Find the equation of the circle.**\r\n\r\n**Solution:**\r\nLet $(x_1, y_1) = (3, 0)$ and $(x_2, y_2) = (0, 4)$.\r\nUsing the diameter form:\r\n$$\r\n\\begin{aligned}\r\n(x - 3)(x - 0) + (y - 0)(y - 4) &= 0 \\\\\r\nx(x - 3) + y(y - 4) &= 0 \\\\\r\nx^2 - 3x + y^2 - 4y &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 3x - 4y = 0$**.\r\n\r\n***\r\n\r\n#### Question 3: Equation from Diameter Endpoints\r\n**Find the equation of the circle having (1, 3) and (2, -5) as the endpoints of a diameter.**\r\n\r\n**Solution:**\r\nLet $(x_1, y_1) = (1, 3)$ and $(x_2, y_2) = (2, -5)$.\r\nUsing the diameter form:\r\n$$\r\n\\begin{aligned}\r\n(x - 1)(x - 2) + (y - 3)(y - (-5)) &= 0 \\\\\r\n(x^2 - 2x - x + 2) + (y - 3)(y + 5) &= 0 \\\\\r\nx^2 - 3x + 2 + y^2 + 5y - 3y - 15 &= 0 \\\\\r\nx^2 + y^2 - 3x + 2y - 13 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 3x + 2y - 13 = 0$**.\r\n\r\n***\r\n\r\n#### Question 4: Equation from Diameter Endpoints\r\n**Find the equation of the circle having the endpoints of a diameter A(5, 6) and B(3, 4).**\r\n\r\n**Solution:**\r\nLet $(x_1, y_1) = (5, 6)$ and $(x_2, y_2) = (3, 4)$.\r\nUsing the diameter form:\r\n$$\r\n\\begin{aligned}\r\n(x - 5)(x - 3) + (y - 6)(y - 4) &= 0 \\\\\r\n(x^2 - 3x - 5x + 15) + (y^2 - 4y - 6y + 24) &= 0 \\\\\r\nx^2 - 8x + 15 + y^2 - 10y + 24 &= 0 \\\\\r\nx^2 + y^2 - 8x - 10y + 39 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 8x - 10y + 39 = 0$**.\r\n\r\n***\r\n\r\n#### Question 5: Equation from Diameter Endpoints\r\n**Find the equation of the circle whose endpoints of a diameter are A(3, 4) and B(2, -7).**\r\n\r\n**Solution:**\r\nLet $(x_1, y_1) = (3, 4)$ and $(x_2, y_2) = (2, -7)$.\r\nUsing the diameter form:\r\n$$\r\n\\begin{aligned}\r\n(x - 3)(x - 2) + (y - 4)(y - (-7)) &= 0 \\\\\r\n(x^2 - 2x - 3x + 6) + (y - 4)(y + 7) &= 0 \\\\\r\nx^2 - 5x + 6 + y^2 + 7y - 4y - 28 &= 0 \\\\\r\nx^2 + y^2 - 5x + 3y - 22 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 5x + 3y - 22 = 0$**.\r\n\r\n***\r\n\r\n#### Question 6: Center from Intersecting Diameters\r\n**Find the coordinates of the center of a circle having equations of two diameters $x + y = 5$ and $2x - y = 1$.**\r\n\r\n**Solution:**\r\nThe center of a circle is the point of intersection of its diameters. We solve the system of linear equations.\r\n$$\r\n\\begin{aligned}\r\nx + y &= 5 \\quad &(1) \\\\\r\n2x - y &= 1 \\quad &(2)\r\n\\end{aligned}\r\n$$\r\nAdding equation (1) and (2):\r\n$$\r\n\\begin{aligned}\r\n(x+y) + (2x-y) &= 5 + 1 \\\\\r\n3x &= 6 \\\\\r\nx &= 2\r\n\\end{aligned}\r\n$$\r\nSubstitute $x=2$ into equation (1):\r\n$$\r\n2 + y = 5 \\implies y = 3\r\n$$\r\nThe center of the circle is **(2, 3)**.\r\n\r\n***\r\n\r\n#### Question 7: Center and Radius from General Equation\r\n**Find the center and radius of a circle having the equation $x^2 + y^2 - 10x - 4y - 7 = 0$.**\r\n\r\n**Solution:**\r\nCompare the given equation to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = -10x &\\implies g = -5 \\\\\r\n2fy = -4y &\\implies f = -2 \\\\\r\nc &= -7\r\n\\end{aligned}\r\n$$\r\n**Center:** $(-g, -f) = (-(-5), -(-2)) = (5, 2)$.\r\n**Radius:** $r = \\sqrt{g^2 + f^2 - c} = \\sqrt{(-5)^2 + (-2)^2 - (-7)} = \\sqrt{25 + 4 + 7} = \\sqrt{36} = 6$.\r\n\r\nThe center is **(5, 2)** and the radius is **6 units**.\r\n\r\n***\r\n\r\n#### Question 8: Finding an Unknown from Radius\r\n**If the radius of the circle $x^2 + y^2 - 4x - 6y - k = 0$ is 4 units, find the value of k.**\r\n\r\n**Solution:**\r\nCompare the equation to $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = -4x &\\implies g = -2 \\\\\r\n2fy = -6y &\\implies f = -3 \\\\\r\nc &= -k\r\n\\end{aligned}\r\n$$\r\nUsing the radius formula, with $r=4$:\r\n$$\r\n\\begin{aligned}\r\nr &= \\sqrt{g^2 + f^2 - c} \\\\\r\n4 &= \\sqrt{(-2)^2 + (-3)^2 - (-k)} \\\\\r\n4 &= \\sqrt{4 + 9 + k} \\\\\r\n4 &= \\sqrt{13 + k}\r\n\\end{aligned}\r\n$$\r\nSquaring both sides:\r\n$$\r\n16 = 13 + k \\implies k = 3\r\n$$\r\nThus, the value of **$k = 3$**.\r\n\r\n***\r\n\r\n#### Question 9: Finding Unknowns from Center\r\n**If the center of the circle $x^2 + y^2 - ax - by - 12 = 0$ is (2, 3), find the value of a and b.**\r\n\r\n**Solution:**\r\nCompare the equation to $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = -ax &\\implies g = -a/2 \\\\\r\n2fy = -by &\\implies f = -b/2\r\n\\end{aligned}\r\n$$\r\nThe center is given by $(-g, -f)$.\r\n$$\r\nCenter = (-(-a/2), -(-b/2)) = (a/2, b/2)\r\n$$\r\nWe are given that the center is (2, 3).\r\n$$\r\n\\begin{aligned}\r\na/2 &= 2 \\implies a = 4 \\\\\r\nb/2 &= 3 \\implies b = 6\r\n\\end{aligned}\r\n$$\r\nThus, **$a=4$ and $b=6$**.\r\n\r\n***\r\n\r\n#### Question 10: Finding an Unknown from Center\r\n**Find the value of m when the center of a circle with equation $x^2 + y^2 - mx + 6y - 12 = 0$ is (-2, -3).**\r\n\r\n**Solution:**\r\nCompare the equation to $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = -mx &\\implies g = -m/2 \\\\\r\n2fy = 6y &\\implies f = 3\r\n\\end{aligned}\r\n$$\r\nThe center is $(-g, -f) = (-(-m/2), -3) = (m/2, -3)$.\r\nWe are given that the center is $(-2, -3)$.\r\nEquating the x-coordinates:\r\n$$\r\n\\frac{m}{2} = -2 \\implies m = -4\r\n$$\r\nThus, **$m = -4$**.\r\n\r\n***\r\n\r\n#### Question 11: Equation from Center and Radius\r\n**Find the equation of a circle having centre (1, -2) and diameter $2\\sqrt{5}$ units.**\r\n\r\n**Solution:**\r\nCenter $(h, k) = (1, -2)$.\r\nDiameter is $2\\sqrt{5}$, so the radius is $r = \\sqrt{5}$.\r\nUsing the standard form $(x-h)^2 + (y-k)^2 = r^2$:\r\n$$\r\n\\begin{aligned}\r\n(x - 1)^2 + (y - (-2))^2 &= (\\sqrt{5})^2 \\\\\r\n(x^2 - 2x + 1) + (y^2 + 4y + 4) &= 5 \\\\\r\nx^2 + y^2 - 2x + 4y + 5 &= 5\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 2x + 4y = 0$**.\r\n\r\n***\r\n\r\n#### Question 12: Equation from Center and Radius\r\n**Find the equation of a circle having centre (2, 3) and radius 5 units.**\r\n\r\n**Solution:**\r\nCenter $(h, k) = (2, 3)$ and radius $r = 5$.\r\nUsing the standard form:\r\n$$\r\n\\begin{aligned}\r\n(x - 2)^2 + (y - 3)^2 &= 5^2 \\\\\r\n(x^2 - 4x + 4) + (y^2 - 6y + 9) &= 25 \\\\\r\nx^2 + y^2 - 4x - 6y + 13 &= 25\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 - 4x - 6y - 12 = 0$**.\r\n\r\n***\r\n\r\n#### Question 13: Center from General Equation\r\n**Find the coordinates of the center of a circle having an equation $x^2 + y^2 + 4x - 5 = 0$.**\r\n\r\n**Solution:**\r\nCompare with $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = 4x &\\implies g = 2 \\\\\r\n2fy = 0y &\\implies f = 0\r\n\\end{aligned}\r\n$$\r\nThe center is $(-g, -f) = (-2, 0)$.\r\nThe center is **(-2, 0)**.\r\n\r\n***\r\n\r\n#### Question 14: Equation from Center and Radius\r\n**Find the equation of a circle having the centre (-1, -3) and radius 3 units.**\r\n\r\n**Solution:**\r\nCenter $(h, k) = (-1, -3)$ and radius $r=3$.\r\nUsing the standard form:\r\n$$\r\n\\begin{aligned}\r\n(x - (-1))^2 + (y - (-3))^2 &= 3^2 \\\\\r\n(x+1)^2 + (y+3)^2 &= 9 \\\\\r\n(x^2 + 2x + 1) + (y^2 + 6y + 9) &= 9 \\\\\r\nx^2 + y^2 + 2x + 6y + 1 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 + 2x + 6y + 1 = 0$**.\r\n\r\n***\r\n\r\n#### Question 15: Center from General Equation\r\n**Find the center of the circle whose equation is $x^2 + y^2 - 6x + 2y + 1 = 0$.**\r\n\r\n**Solution:**\r\nCompare with $x^2 + y^2 + 2gx + 2fy + c = 0$.\r\n$$\r\n\\begin{aligned}\r\n2gx = -6x &\\implies g = -3 \\\\\r\n2fy = 2y &\\implies f = 1\r\n\\end{aligned}\r\n$$\r\nThe center is $(-g, -f) = (-(-3), -1) = (3, -1)$.\r\nThe center is **(3, -1)**.\r\n\r\n---\r\n\r\n### Long Questions\r\n\r\n***\r\n\r\n#### Question 16: Equation of Circle Passing Through Center of Another\r\n**Circle A with centre X(-4, 5) passes through the centre Y of circle B. If the equation of circle B is $x^2 + y^2 - 4x + 6y - 12 = 0$, find the equation of circle A.**\r\n\r\n**Solution:**\r\n**Step 1: Find the center of circle B.**\r\nCompare $x^2 + y^2 - 4x + 6y - 12 = 0$ with the general form.\r\n$g = -2$, $f=3$.\r\nCenter of B is $Y = (-g, -f) = (2, -3)$.\r\n\r\n**Step 2: Find the radius of circle A.**\r\nThe radius of circle A is the distance between its center X(-4, 5) and the point it passes through, Y(2, -3).\r\n$$\r\n\\begin{aligned}\r\nr^2 &= (x_2 - x_1)^2 + (y_2 - y_1)^2 \\\\\r\n&= (2 - (-4))^2 + (-3 - 5)^2 \\\\\r\n&= (6)^2 + (-8)^2 \\\\\r\n&= 36 + 64 = 100\r\n\\end{aligned}\r\n$$\r\n**Step 3: Find the equation of circle A.**\r\nThe center of circle A is X(h, k) = (-4, 5) and $r^2=100$.\r\n$$\r\n\\begin{aligned}\r\n(x - (-4))^2 + (y - 5)^2 &= 100 \\\\\r\n(x+4)^2 + (y-5)^2 &= 100 \\\\\r\nx^2 + 8x + 16 + y^2 - 10y + 25 &= 100 \\\\\r\nx^2 + y^2 + 8x - 10y - 59 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 + 8x - 10y - 59 = 0$**.\r\n\r\n***\r\n\r\n#### Question 17: Equation of Circle Passing Through Center of Another\r\n**Find the equation of a circle with centre (-1, 2) and passing through the centre of the circle having equation $x^2 + y^2 - 6x - 10y - 2 = 0$.**\r\n\r\n**Solution:**\r\n**Step 1: Find the center of the given circle.**\r\nComparing $x^2 + y^2 - 6x - 10y - 2 = 0$ to the general form.\r\n$g=-3$, $f=-5$.\r\nThe center is $(-g, -f) = (3, 5)$.\r\n\r\n**Step 2: Find the radius of the required circle.**\r\nThe required circle has center $(h,k)=(-1,2)$ and passes through $(3,5)$. The radius is the distance between these two points.\r\n$$\r\n\\begin{aligned}\r\nr^2 &= (3 - (-1))^2 + (5 - 2)^2 \\\\\r\n&= (4)^2 + (3)^2 \\\\\r\n&= 16 + 9 = 25\r\n\\end{aligned}\r\n$$\r\n**Step 3: Find the equation of the required circle.**\r\nCenter is $(-1, 2)$ and $r^2=25$.\r\n$$\r\n\\begin{aligned}\r\n(x - (-1))^2 + (y - 2)^2 &= 25 \\\\\r\n(x+1)^2 + (y-2)^2 &= 25 \\\\\r\nx^2 + 2x + 1 + y^2 - 4y + 4 &= 25 \\\\\r\nx^2 + y^2 + 2x - 4y - 20 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 + 2x - 4y - 20 = 0$**.\r\n\r\n***\r\n\r\n#### Question 18: Finding the Other End of a Diameter\r\n**If one endpoint of a diameter of the circle $x^2 + y^2 + 4x - 6y + 8 = 0$ is (-4, 4), find the coordinates of the other endpoint.**\r\n\r\n**Solution:**\r\n**Step 1: Find the center of the circle.**\r\nComparing the equation to the general form: $g=2$, $f=-3$.\r\nThe center is $(-g, -f) = (-2, 3)$.\r\n\r\n**Step 2: Use the midpoint formula.**\r\nThe center of a circle is the midpoint of its diameter. Let the other endpoint be $(x_2, y_2)$.\r\n$$\r\n\\begin{aligned}\r\n\\text{Center} &= \\left( \\frac{x_1+x_2}{2}, \\frac{y_1+y_2}{2} \\right) \\\\\r\n(-2, 3) &= \\left( \\frac{-4+x_2}{2}, \\frac{4+y_2}{2} \\right)\r\n\\end{aligned}\r\n$$\r\nEquating coordinates:\r\n$$\r\n-2 = \\frac{-4+x_2}{2} \\implies -4 = -4+x_2 \\implies x_2=0\r\n$$\r\n$$\r\n3 = \\frac{4+y_2}{2} \\implies 6 = 4+y_2 \\implies y_2=2\r\n$$\r\nThe coordinates of the other end are **(0, 2)**.\r\n\r\n***\r\n\r\n#### Question 19: Equation of a Concentric Circle\r\n**Find the equation of a circle concentric with the circle $2x^2 + 2y^2 + 4x - 2y + 1 = 0$ and passing through the point (4,-2).**\r\n\r\n**Solution:**\r\n**Step 1: Find the center of the given circle.**\r\nFirst, divide the equation by 2 to get the standard general form:\r\n$x^2 + y^2 + 2x - y + \\frac{1}{2} = 0$.\r\nComparing with the general form, $g=1$ and $f=-1/2$.\r\nThe center is $(-g, -f) = (-1, 1/2)$.\r\n\r\n**Step 2: Find the radius of the required circle.**\r\nSince the required circle is concentric, its center is also $(-1, 1/2)$. It passes through $(4, -2)$. The radius is the distance between these points.\r\n$$\r\n\\begin{aligned}\r\nr^2 &= (4 - (-1))^2 + (-2 - \\frac{1}{2})^2 \\\\\r\n&= (5)^2 + (-\\frac{5}{2})^2 \\\\\r\n&= 25 + \\frac{25}{4} = \\frac{100+25}{4} = \\frac{125}{4}\r\n\\end{aligned}\r\n$$\r\n**Step 3: Write the equation of the required circle.**\r\nCenter is $(-1, 1/2)$ and $r^2 = 125/4$.\r\n$$\r\n\\begin{aligned}\r\n(x - (-1))^2 + (y - \\frac{1}{2})^2 &= \\frac{125}{4} \\\\\r\n(x+1)^2 + (y - \\frac{1}{2})^2 &= \\frac{125}{4} \\\\\r\nx^2+2x+1 + y^2 - y + \\frac{1}{4} &= \\frac{125}{4}\r\n\\end{aligned}\r\n$$\r\nMultiply by 4 to clear the fraction:\r\n$$\r\n\\begin{aligned}\r\n4x^2 + 8x + 4 + 4y^2 - 4y + 1 &= 125 \\\\\r\n4x^2 + 4y^2 + 8x - 4y - 120 &= 0 \\\\\r\nx^2 + y^2 + 2x - y - 30 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 + 2x - y - 30 = 0$**.\r\n\r\n***\r\n\r\n#### Question 20: Equation of a Concentric Circle\r\n**Find the equation of a circle passing through the point (4, 3) and concentric with the circle having equation $x^2 + y^2 + 6x - 8y - 11 = 0$.**\r\n\r\n**Solution:**\r\n**Step 1: Find the center of the given circle.**\r\nCompare $x^2 + y^2 + 6x - 8y - 11 = 0$ with the general form.\r\n$g=3$, $f=-4$. The center is $(-g,-f) = (-3, 4)$.\r\n\r\n**Step 2: Find the radius of the required circle.**\r\nSince the new circle is concentric, its center is also $(-3, 4)$. It passes through the point $(4, 3)$. The radius is the distance between these points.\r\n$$\r\n\\begin{aligned}\r\nr^2 &= (4 - (-3))^2 + (3 - 4)^2 \\\\\r\n&= (7)^2 + (-1)^2 \\\\\r\n&= 49 + 1 = 50\r\n\\end{aligned}\r\n$$\r\n**Step 3: Write the equation of the required circle.**\r\nCenter is $(-3, 4)$ and $r^2 = 50$.\r\n$$\r\n\\begin{aligned}\r\n(x - (-3))^2 + (y - 4)^2 &= 50 \\\\\r\n(x+3)^2 + (y-4)^2 &= 50 \\\\\r\nx^2+6x+9 + y^2-8y+16 &= 50 \\\\\r\nx^2+y^2+6x-8y-25 &= 0\r\n\\end{aligned}\r\n$$\r\nThe required equation is **$x^2 + y^2 + 6x - 8y - 25 = 0$**.\r\n\r\n***\r\n\r\n#### Question 21: Equation of a Circle from Two Points and Center on a Line\r\n**Find the equation of the circle whose centre lies on the line $x - 4y = 1$ and which passes through the points (3, 7) and (5, 5).**\r\n\r\n**Solution:**\r\nLet the center be $(h,k)$. Since it lies on the line $x - 4y = 1$, we have $h - 4k = 1 \\implies h = 4k+1$.\r\nThe distance from the center to any point on the circle is the radius. So, the squared distance from $(h,k)$ to $(3,7)$ is the same as to $(5,5)$.\r\n$$\r\n\\begin{aligned}\r\n(h-3)^2 + (k-7)^2 &= (h-5)^2 + (k-5)^2 \\\\\r\nh^2 - 6h + 9 + k^2 - 14k + 49 &= h^2 - 10h + 25 + k^2 - 10k + 25 \\\\\r\n-6h - 14k + 58 &= -10h - 10k + 50 \\\\\r\n4h - 4k &= -8 \\\\\r\nh - k &= -2\r\n\\end{aligned}\r\n$$\r\nNow substitute $h = 4k+1$ into this new equation:\r\n$$\r\n\\begin{aligned}\r\n(4k+1) - k &= -2 \\\\\r\n3k &= -3 \\\\\r\nk &= -1\r\n\\end{aligned}\r\n$$\r\nNow find h: $h = 4(-1)+1 = -3$. The center is $(-3,-1)$.\r\nFind the radius squared, $r^2$, using point $(3,7)$:\r\n$$\r\nr^2 = (3 - (-3))^2 + (7 - (-1))^2 = 6^2 + 8^2 = 36+64=100\r\n$$\r\nThe equation is $(x+3)^2 + (y+1)^2 = 100$, which expands to **$x^2 + y^2 + 6x + 2y - 90 = 0$**.\r\n\r\n***\r\n\r\n#### Question 22: Equation of a Circle from Three Points\r\n**Find the equation of the circle which passes through the points P(5, 7), Q(6, 6) and R(2, -2).**\r\n\r\n**Solution:**\r\nLet the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$.\r\nThe squared distance from the center $(h,k)$ to each point is equal.\r\n**From P and Q:**\r\n$$\r\n\\begin{aligned}\r\n(h-5)^2 + (k-7)^2 &= (h-6)^2 + (k-6)^2 \\\\\r\nh^2-10h+25 + k^2-14k+49 &= h^2-12h+36 + k^2-12k+36 \\\\\r\n-10h-14k+74 &= -12h-12k+72 \\\\\r\n2h - 2k + 2 &= 0 \\implies h - k + 1 = 0 \\implies h = k-1 \\quad \\text{(Eq. 1)}\r\n\\end{aligned}\r\n$$\r\n**From Q and R:**\r\n$$\r\n\\begin{aligned}\r\n(h-6)^2 + (k-6)^2 &= (h-2)^2 + (k-(-2))^2 \\\\\r\nh^2-12h+36 + k^2-12k+36 &= h^2-4h+4 + k^2+4k+4 \\\\\r\n-12h-12k+72 &= -4h+4k+8 \\\\\r\n64 &= 8h + 16k \\\\\r\n8 &= h + 2k \\quad \\text{(Eq. 2)}\r\n\\end{aligned}\r\n$$\r\n**Solve for h and k:**\r\nSubstitute Eq. 1 into Eq. 2:\r\n$8 = (k-1) + 2k \\implies 9 = 3k \\implies k=3$.\r\nThen, $h = 3-1=2$. The center is $(h,k)=(2,3)$.\r\n\r\n**Find $r^2$ using point P(5,7):**\r\n$r^2 = (5-2)^2 + (7-3)^2 = 3^2 + 4^2 = 9+16=25$.\r\n\r\n**Write the equation:**\r\n$$(x-2)^2 + (y-3)^2 = 25$$\r\nThe required equation is **$x^2 + y^2 - 4x - 6y - 12 = 0$**.', 1, '2025-07-25 01:11:38', '2025-09-18 07:21:49'),
(145, 14241, '# Grade 10 Optional Mathematics: Scalar (Dot) Product\r\n## Unit: Vector | Chapter: Scalar Product | Complete Solutions\r\n\r\n---\r\n\r\n## 🎯 Chapter Introduction\r\n\r\nThis guide offers a complete overview of the **Scalar Product** (also known as the Dot Product) of vectors. It starts with a clear breakdown of the essential concepts and formulas, then provides detailed, step-by-step solutions to a wide range of problems, including finding the angle between vectors, determining vector magnitudes, and applying the condition for perpendicularity. 📐\r\n\r\n### Key Learning Areas\r\n1.  **Definition of Scalar Product:** Understanding what the scalar product is and how to calculate it.\r\n2.  **Magnitude of a Vector:** Calculating the length or magnitude of a vector.\r\n3.  **Angle Between Two Vectors:** Using the scalar product formula to find the angle between two vectors.\r\n4.  **Condition for Perpendicularity:** Using the scalar product to prove that two vectors are perpendicular.\r\n5.  **Properties of Unit Vectors:** Understanding the scalar products of unit vectors $\\overrightarrow{i}$ and $\\overrightarrow{j}$.\r\n\r\n---\r\n\r\n## 🔑 Key Concepts & Formulas\r\n\r\nBefore diving into the problems, let\'s review the fundamental concepts.\r\n\r\n### 1. The Scalar (Dot) Product\r\n\r\nThe scalar product of two vectors, $\\overrightarrow{a}$ and $\\overrightarrow{b}$, is a scalar (a number, not a vector) quantity.\r\n\r\n* **Geometric Formula:**\r\n    $$\\overrightarrow{a} \\cdot \\overrightarrow{b} = |\\overrightarrow{a}| |\\overrightarrow{b}| \\cos\\theta$$\r\n    Where:\r\n    * $|\\overrightarrow{a}|$ is the magnitude (length) of vector $\\overrightarrow{a}$.\r\n    * $|\\overrightarrow{b}|$ is the magnitude (length) of vector $\\overrightarrow{b}$.\r\n    * $\\theta$ is the angle between the two vectors.\r\n\r\n\r\n\r\n* **Component Formula:**\r\n    If $\\overrightarrow{a} = x_1\\overrightarrow{i} + y_1\\overrightarrow{j}$ and $\\overrightarrow{b} = x_2\\overrightarrow{i} + y_2\\overrightarrow{j}$, then:\r\n    $$\\overrightarrow{a} \\cdot \\overrightarrow{b} = x_1x_2 + y_1y_2$$\r\n\r\n---\r\n\r\n### 2. Magnitude of a Vector\r\n\r\nThe magnitude of a vector $\\overrightarrow{a} = x\\overrightarrow{i} + y\\overrightarrow{j}$ is its length.\r\n\r\n* **Formula:**\r\n    $$|\\overrightarrow{a}| = \\sqrt{x^2 + y^2}$$\r\n\r\n---\r\n\r\n### 3. Angle Between Two Vectors\r\n\r\nWe can rearrange the geometric formula to solve for the angle $\\theta$.\r\n\r\n* **Formula:**\r\n    $$\\cos\\theta = \\frac{\\overrightarrow{a} \\cdot \\overrightarrow{b}}{|\\overrightarrow{a}| |\\overrightarrow{b}|}$$\r\n\r\n---\r\n\r\n### 4. Condition for Perpendicular Vectors\r\n\r\nTwo non-zero vectors are perpendicular (orthogonal) if the angle between them is 90°. Since $\\cos(90^\\circ) = 0$, the condition is:\r\n\r\n* **Condition:**\r\n    $$\\overrightarrow{a} \\cdot \\overrightarrow{b} = 0$$\r\n\r\n---\r\n\r\n### 5. Properties of Unit Vectors $\\overrightarrow{i}$ and $\\overrightarrow{j}$\r\n\r\n* The scalar product of a unit vector with itself is 1.\r\n    $$\\overrightarrow{i} \\cdot \\overrightarrow{i} = 1 \\quad \\text{and} \\quad \\overrightarrow{j} \\cdot \\overrightarrow{j} = 1$$\r\n* The scalar product of two different standard unit vectors is 0 (as they are perpendicular).\r\n    $$\\overrightarrow{i} \\cdot \\overrightarrow{j} = 0 \\quad \\text{and} \\quad \\overrightarrow{j} \\cdot \\overrightarrow{i} = 0$$\r\n* A vector dotted with itself equals the square of its magnitude:\r\n    $$\\overrightarrow{a} \\cdot \\overrightarrow{a} = |\\overrightarrow{a}|^2$$\r\n\r\n---\r\n\r\n## 📚 Exercise Solutions\r\n\r\n### Short Questions\r\n\r\n***\r\n\r\n#### Question 1: Finding Magnitude of a Vector\r\n**If $\\overrightarrow{p} \\cdot \\overrightarrow{q} = 6$, $|\\overrightarrow{q}| = 2\\sqrt{3}$ and the angle between $\\overrightarrow{p}$ and $\\overrightarrow{q}$ is 30°, find the length of $\\overrightarrow{p}$.**\r\n\r\n**Solution:**\r\nGiven: $\\overrightarrow{p} \\cdot \\overrightarrow{q} = 6$, $|\\overrightarrow{q}| = 2\\sqrt{3}$, and $\\theta = 30^\\circ$.\r\nWe know the scalar product formula:\r\n$$ \\overrightarrow{p} \\cdot \\overrightarrow{q} = |\\overrightarrow{p}| |\\overrightarrow{q}| \\cos\\theta $$\r\nSubstituting the given values:\r\n$$\r\n\\begin{aligned}\r\n6 &= |\\overrightarrow{p}| \\cdot 2\\sqrt{3} \\cdot \\cos(30^\\circ) \\\\\r\n6 &= |\\overrightarrow{p}| \\cdot 2\\sqrt{3} \\cdot \\frac{\\sqrt{3}}{2} \\\\\r\n6 &= |\\overrightarrow{p}| \\cdot 3 \\\\\r\n|\\overrightarrow{p}| &= \\frac{6}{3} = 2\r\n\\end{aligned}\r\n$$\r\nThe length of $\\overrightarrow{p}$ is **2 units**.\r\n\r\n***\r\n\r\n#### Question 2: Finding the Angle Between Two Vectors\r\n**If $|\\overrightarrow{a}| = 4$, $|\\overrightarrow{b}| = 5$ and $\\overrightarrow{a} \\cdot \\overrightarrow{b} = 10$, then find the angle between $\\overrightarrow{a}$ and $\\overrightarrow{b}$.**\r\n\r\n**Solution:**\r\nUsing the formula for the angle between two vectors:\r\n$$\r\n\\begin{aligned}\r\n\\cos\\theta &= \\frac{\\overrightarrow{a} \\cdot \\overrightarrow{b}}{|\\overrightarrow{a}| |\\overrightarrow{b}|} \\\\\r\n\\cos\\theta &= \\frac{10}{4 \\cdot 5} \\\\\r\n\\cos\\theta &= \\frac{10}{20} = \\frac{1}{2}\r\n\\end{aligned}\r\n$$\r\nSince $\\cos(60^\\circ) = \\frac{1}{2}$, the angle is **$\\theta = 60^\\circ$**.\r\n\r\n***\r\n\r\n#### Question 3: Finding the Scalar Product\r\n**If $|\\overrightarrow{a}| = 3\\sqrt{3}$, $|\\overrightarrow{b}| = 4$ and angle between $\\overrightarrow{a}$ and $\\overrightarrow{b}$ is 60°, then find the value of $\\overrightarrow{a} \\cdot \\overrightarrow{b}$.**\r\n\r\n**Solution:**\r\nUsing the scalar product formula:\r\n$$\r\n\\begin{aligned}\r\n\\overrightarrow{a} \\cdot \\overrightarrow{b} &= |\\overrightarrow{a}| |\\overrightarrow{b}| \\cos\\theta \\\\\r\n&= 3\\sqrt{3} \\cdot 4 \\cdot \\cos(60^\\circ) \\\\\r\n&= 12\\sqrt{3} \\cdot \\frac{1}{2} \\\\\r\n&= 6\\sqrt{3}\r\n\\end{aligned}\r\n$$\r\nThe value of $\\overrightarrow{a} \\cdot \\overrightarrow{b}$ is **$6\\sqrt{3}$**.\r\n\r\n***\r\n\r\n#### Question 4: Finding the Scalar Product\r\n**If $|\\overrightarrow{p}| = 6$, $|\\overrightarrow{q}| = 6$ and the angle between them is 60°, find the value of $\\overrightarrow{p} \\cdot \\overrightarrow{q}$.**\r\n\r\n**Solution:**\r\nUsing the scalar product formula:\r\n$$\r\n\\begin{aligned}\r\n\\overrightarrow{p} \\cdot \\overrightarrow{q} &= |\\overrightarrow{p}| |\\overrightarrow{q}| \\cos\\theta \\\\\r\n&= 6 \\cdot 6 \\cdot \\cos(60^\\circ) \\\\\r\n&= 36 \\cdot \\frac{1}{2} \\\\\r\n&= 18\r\n\\end{aligned}\r\n$$\r\nThe value is **18**.\r\n\r\n***\r\n\r\n#### Question 5: Finding an Unknown in a Perpendicular Vector\r\n**If the vectors $10\\overrightarrow{i} - 7\\overrightarrow{j}$ and $a\\overrightarrow{i} + 10\\overrightarrow{j}$ are perpendicular to each other, find the value of a.**\r\n\r\n**Solution:**\r\nLet $\\overrightarrow{p} = 10\\overrightarrow{i} - 7\\overrightarrow{j}$ and $\\overrightarrow{q} = a\\overrightarrow{i} + 10\\overrightarrow{j}$.\r\nFor two vectors to be perpendicular, their scalar product must be zero.\r\n$$\r\n\\begin{aligned}\r\n\\overrightarrow{p} \\cdot \\overrightarrow{q} &= 0 \\\\\r\n(10\\overrightarrow{i} - 7\\overrightarrow{j}) \\cdot (a\\overrightarrow{i} + 10\\overrightarrow{j}) &= 0 \\\\\r\n(10 \\cdot a) + (-7 \\cdot 10) &= 0 \\\\\r\n10a - 70 &= 0 \\\\\r\n10a &= 70 \\\\\r\na &= 7\r\n\\end{aligned}\r\n$$\r\nThe value of **$a=7$**.\r\n\r\n***\r\n\r\n#### Question 6: Finding an Unknown in a Perpendicular Vector\r\n**If $\\overrightarrow{a} = -5\\overrightarrow{i} + 3\\overrightarrow{j}$ and $\\overrightarrow{b} = p\\overrightarrow{i} + (p+2)\\overrightarrow{j}$ are perpendicular to each other, find the value of p.**\r\n\r\n**Solution:**\r\nFor perpendicular vectors, $\\overrightarrow{a} \\cdot \\overrightarrow{b} = 0$.\r\n$$\r\n\\begin{aligned}\r\n(-5\\overrightarrow{i} + 3\\overrightarrow{j}) \\cdot (p\\overrightarrow{i} + (p+2)\\overrightarrow{j}) &= 0 \\\\\r\n(-5 \\cdot p) + (3 \\cdot (p+2)) &= 0 \\\\\r\n-5p + 3p + 6 &= 0 \\\\\r\n-2p &= -6 \\\\\r\np &= 3\r\n\\end{aligned}\r\n$$\r\nThe value of **$p=3$**.\r\n\r\n***\r\n\r\n#### Question 7: Finding Magnitude of a Vector\r\n**If $|\\overrightarrow{b}| = 6$, $\\overrightarrow{a} \\cdot \\overrightarrow{b} = 12$ and the angle between $\\overrightarrow{a}$ and $\\overrightarrow{b}$ is 60°, find the value of $|\\overrightarrow{a}|$.**\r\n\r\n**Solution:**\r\nUsing the scalar product formula:\r\n$$\r\n\\begin{aligned}\r\n\\overrightarrow{a} \\cdot \\overrightarrow{b} &= |\\overrightarrow{a}| |\\overrightarrow{b}| \\cos\\theta \\\\\r\n12 &= |\\overrightarrow{a}| \\cdot 6 \\cdot \\cos(60^\\circ) \\\\\r\n12 &= |\\overrightarrow{a}| \\cdot 6 \\cdot \\frac{1}{2} \\\\\r\n12 &= |\\overrightarrow{a}| \\cdot 3 \\\\\r\n|\\overrightarrow{a}| &= 4\r\n\\end{aligned}\r\n$$\r\nThe magnitude of $\\overrightarrow{a}$ is **4 units**.\r\n\r\n***\r\n\r\n#### Question 8: Finding the Angle Between Two Vectors\r\n**Find the angle between the unit vector $\\overrightarrow{i}$ and the vector $\\overrightarrow{a} = \\sqrt{3}\\overrightarrow{i} + \\overrightarrow{j}$.**\r\n\r\n**Solution:**\r\nLet the first vector be $\\overrightarrow{u} = \\overrightarrow{i} = 1\\overrightarrow{i} + 0\\overrightarrow{j}$.\r\nLet the second vector be $\\overrightarrow{a} = \\sqrt{3}\\overrightarrow{i} + 1\\overrightarrow{j}$.\r\n\r\n**Step 1: Find the scalar product.**\r\n$$ \\overrightarrow{u} \\cdot \\overrightarrow{a} = (1)(\\sqrt{3}) + (0)(1) = \\sqrt{3} $$\r\n**Step 2: Find the magnitudes.**\r\n$$ |\\overrightarrow{u}| = |\\overrightarrow{i}| = 1 $$\r\n$$ |\\overrightarrow{a}| = \\sqrt{(\\sqrt{3})^2 + 1^2} = \\sqrt{3+1} = \\sqrt{4} = 2 $$\r\n**Step 3: Use the angle formula.**\r\n$$\r\n\\begin{aligned}\r\n\\cos\\theta &= \\frac{\\overrightarrow{u} \\cdot \\overrightarrow{a}}{|\\overrightarrow{u}| |\\overrightarrow{a}|} = \\frac{\\sqrt{3}}{1 \\cdot 2} = \\frac{\\sqrt{3}}{2}\r\n\\end{aligned}\r\n$$\r\nSince $\\cos(30^\\circ) = \\frac{\\sqrt{3}}{2}$, the angle is **$\\theta = 30^\\circ$**.\r\n\r\n---\r\n\r\n### Long Questions\r\n\r\n***\r\n\r\n#### Question 9: Angle Between Unit Vectors\r\n**If $\\overrightarrow{a} + 2\\overrightarrow{b}$ and $5\\overrightarrow{a} - 4\\overrightarrow{b}$ are perpendicular to each other and $\\overrightarrow{a}$ and $\\overrightarrow{b}$ are unit vectors, find the angle between $\\overrightarrow{a}$ and $\\overrightarrow{b}$.**\r\n\r\n**Solution:**\r\nSince the two vectors are perpendicular, their scalar product is zero.\r\n$$ (\\overrightarrow{a} + 2\\overrightarrow{b}) \\cdot (5\\overrightarrow{a} - 4\\overrightarrow{b}) = 0 $$\r\nExpanding the dot product:\r\n$$\r\n\\begin{aligned}\r\n5(\\overrightarrow{a} \\cdot \\overrightarrow{a}) - 4(\\overrightarrow{a} \\cdot \\overrightarrow{b}) + 10(\\overrightarrow{b} \\cdot \\overrightarrow{a}) - 8(\\overrightarrow{b} \\cdot \\overrightarrow{b}) &= 0 \\\\\r\n5|\\overrightarrow{a}|^2 + 6(\\overrightarrow{a} \\cdot \\overrightarrow{b}) - 8|\\overrightarrow{b}|^2 &= 0\r\n\\end{aligned}\r\n$$\r\nSince $\\overrightarrow{a}$ and $\\overrightarrow{b}$ are unit vectors, $|\\overrightarrow{a}|=1$ and $|\\overrightarrow{b}|=1$.\r\n$$\r\n\\begin{aligned}\r\n5(1)^2 + 6(\\overrightarrow{a} \\cdot \\overrightarrow{b}) - 8(1)^2 &= 0 \\\\\r\n5 + 6(\\overrightarrow{a} \\cdot \\overrightarrow{b}) - 8 &= 0 \\\\\r\n6(\\overrightarrow{a} \\cdot \\overrightarrow{b}) &= 3 \\\\\r\n\\overrightarrow{a} \\cdot \\overrightarrow{b} &= \\frac{1}{2}\r\n\\end{aligned}\r\n$$\r\nNow, use the angle formula:\r\n$$\r\n\\cos\\theta = \\frac{\\overrightarrow{a} \\cdot \\overrightarrow{b}}{|\\overrightarrow{a}| |\\overrightarrow{b}|} = \\frac{1/2}{1 \\cdot 1} = \\frac{1}{2}\r\n$$\r\nThus, the angle is **$\\theta = 60^\\circ$**.\r\n\r\n***\r\n\r\n#### Question 10: Finding the Angle Between Two Vectors\r\n**If $\\overrightarrow{p} + \\overrightarrow{q} + \\overrightarrow{r} = 0$, $|\\overrightarrow{p}| = 6$, $|\\overrightarrow{q}| = 7$ and $|\\overrightarrow{r}| = \\sqrt{127}$, find the angle between $\\overrightarrow{p}$ and $\\overrightarrow{q}$.**\r\n\r\n**Solution:**\r\nFrom the given condition:\r\n$$ \\overrightarrow{p} + \\overrightarrow{q} = -\\overrightarrow{r} $$\r\nSquaring both sides (taking the dot product with itself):\r\n$$\r\n\\begin{aligned}\r\n(\\overrightarrow{p} + \\overrightarrow{q}) \\cdot (\\overrightarrow{p} + \\overrightarrow{q}) &= (-\\overrightarrow{r}) \\cdot (-\\overrightarrow{r}) \\\\\r\n|\\overrightarrow{p}|^2 + 2(\\overrightarrow{p} \\cdot \\overrightarrow{q}) + |\\overrightarrow{q}|^2 &= |\\overrightarrow{r}|^2\r\n\\end{aligned}\r\n$$\r\nSubstitute the given magnitudes:\r\n$$\r\n\\begin{aligned}\r\n6^2 + 2(\\overrightarrow{p} \\cdot \\overrightarrow{q}) + 7^2 &= (\\sqrt{127})^2 \\\\\r\n36 + 2(\\overrightarrow{p} \\cdot \\overrightarrow{q}) + 49 &= 127 \\\\\r\n85 + 2(\\overrightarrow{p} \\cdot \\overrightarrow{q}) &= 127 \\\\\r\n2(\\overrightarrow{p} \\cdot \\overrightarrow{q}) &= 42 \\\\\r\n\\overrightarrow{p} \\cdot \\overrightarrow{q} &= 21\r\n\\end{aligned}\r\n$$\r\nNow, find the angle $\\theta$:\r\n$$\r\n\\cos\\theta = \\frac{\\overrightarrow{p} \\cdot \\overrightarrow{q}}{|\\overrightarrow{p}| |\\overrightarrow{q}|} = \\frac{21}{6 \\cdot 7} = \\frac{21}{42} = \\frac{1}{2}\r\n$$\r\nThus, the angle is **$\\theta = 60^\\circ$**.', 1, '2025-07-25 01:11:38', '2025-09-18 09:23:05');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(148, 14240, '# Grade 10 Optional Mathematics: Vector Geometry\r\n## Unit: Vector | Chapter: Geometric Proofs & Section Formula | Complete Solutions\r\n\r\n---\r\n\r\n## 🎯 Key Concepts & Formulas\r\n\r\nThis section covers the fundamental principles of vector geometry, including addition laws, position vectors, and formulas for dividing a line segment. These concepts are the building blocks for the proofs and problems that follow.\r\n\r\n### 1. Vector Addition\r\n\r\n* **Triangle Law of Vector Addition:** If two vectors are represented by two sides of a triangle in sequence, then the third side of the triangle, taken in the opposite order, represents the sum (resultant) of the two vectors. For $\\triangle ABC$, $\\overrightarrow{AB} + \\overrightarrow{BC} = \\overrightarrow{AC}$.\r\n* **Parallelogram Law of Vector Addition:** If two vectors acting simultaneously from a point are represented by the adjacent sides of a parallelogram, their resultant is represented by the diagonal of the parallelogram drawn from the same point. For parallelogram OABC with origin O, $\\overrightarrow{OA} + \\overrightarrow{OC} = \\overrightarrow{OB}$.\r\n\r\n### 2. Position Vector\r\n\r\nThe position vector of a point A, denoted as $\\overrightarrow{OA}$ or $\\overrightarrow{a}$, is the vector from a fixed origin (O) to that point. It\'s used to define the location of a point in space. The vector from point A to point B can be expressed using position vectors as $\\overrightarrow{AB} = \\overrightarrow{OB} - \\overrightarrow{OA}$.\r\n\r\n### 3. Section Formula\r\n\r\nThis formula finds the position vector of a point that divides the line segment joining two points in a specific ratio.\r\n\r\n* **Internal Division:** If a point M divides the line segment joining P (position vector $\\overrightarrow{p}$) and Q (position vector $\\overrightarrow{q}$) internally in the ratio $m_1:m_2$, the position vector of M is:\r\n    $$ \\overrightarrow{OM} = \\frac{m_1\\overrightarrow{q} + m_2\\overrightarrow{p}}{m_1 + m_2} $$\r\n* **External Division:** If M divides the line joining A ($\\overrightarrow{a}$) and B ($\\overrightarrow{b}$) externally in the ratio $m:n$, its position vector is:\r\n    $$ \\overrightarrow{OM} = \\frac{m\\overrightarrow{b} - n\\overrightarrow{a}}{m - n} $$\r\n\r\n### 4. Mid-Point Formula\r\n\r\nThis is a special case of the internal section formula where the ratio is 1:1.\r\n\r\n* **Formula:** The position vector of the mid-point M of the line segment joining points A and B is:\r\n    $$ \\overrightarrow{OM} = \\frac{\\overrightarrow{OA} + \\overrightarrow{OB}}{2} $$\r\n\r\n### 5. Centroid Formula\r\n\r\nThe centroid is the point where the three medians of a triangle intersect.\r\n\r\n* **Formula:** The position vector of the centroid G of a triangle ABC is the average of the position vectors of its vertices:\r\n    $$ \\overrightarrow{OG} = \\frac{\\overrightarrow{OA} + \\overrightarrow{OB} + \\overrightarrow{OC}}{3} $$\r\n\r\n### 6. Scalar (Dot) Product and Perpendicularity\r\n\r\n* **Condition for Perpendicularity:** Two non-zero vectors $\\overrightarrow{a}$ and $\\overrightarrow{b}$ are perpendicular (orthogonal) if and only if their scalar product is zero.\r\n    $$ \\overrightarrow{a} \\cdot \\overrightarrow{b} = 0 $$\r\n\r\n---\r\n\r\n## 📐 Vector Proofs of Geometric Theorems\r\n\r\nThis section provides step-by-step proofs for key geometric theorems using vector methods.\r\n\r\n***\r\n\r\n### 1. Angle in a Semi-circle is a Right Angle (Method 1)\r\n\r\n**Prove by vector method that the angle at the circumference of a semi-circle is a right angle.**\r\n\r\n**Given:**\r\n* ACB is a semi-circle with center O.\r\n* AB is the diameter.\r\n\r\n\r\n\r\n**To Prove:** $\\angle ACB = 90^\\circ$\r\n\r\n**Construction:** Join O to C.\r\n\r\n**Proof:**\r\n\r\n| Statement | Reason |\r\n| :--- | :--- |\r\n| i) $\\overrightarrow{AC} = \\overrightarrow{AO} + \\overrightarrow{OC}$ | By Triangle Law of Vector Addition. |\r\n| ii) $\\overrightarrow{BC} = \\overrightarrow{BO} + \\overrightarrow{OC}$ | By Triangle Law of Vector Addition. |\r\n| iii) $\\overrightarrow{BC} = -\\overrightarrow{OA} + \\overrightarrow{OC}$ | $\\overrightarrow{BO} = -\\overrightarrow{OA}$ since O is the center and OA and OB are radii in opposite directions. |\r\n| iv) $\\overrightarrow{AC} \\cdot \\overrightarrow{BC} = (\\overrightarrow{OC} + \\overrightarrow{OA}) \\cdot (\\overrightarrow{OC} - \\overrightarrow{OA})$ | Taking the dot product of the rearranged vectors from (i) and (iii). |\r\n| v) $\\overrightarrow{AC} \\cdot \\overrightarrow{BC} = |\\overrightarrow{OC}|^2 - |\\overrightarrow{OA}|^2$ | Using the algebraic formula $(a+b)(a-b) = a^2 - b^2$. |\r\n| vi) $\\overrightarrow{AC} \\cdot \\overrightarrow{BC} = |\\overrightarrow{OC}|^2 - |\\overrightarrow{OC}|^2 = 0$ | $|\\overrightarrow{OA}| = |\\overrightarrow{OC}|$ (radii of the same circle). |\r\n| vii) $\\overrightarrow{AC} \\perp \\overrightarrow{BC}$ | The scalar (dot) product of two non-zero vectors is zero. |\r\n| viii) $\\angle ACB = 90^\\circ$ | From statement (vii), the angle between the vectors is 90°. |\r\n\r\n**Hence, the angle in a semi-circle is a right angle.**\r\n\r\n***\r\n\r\n### 2. Angle in a Semi-circle is a Right Angle (Method 2)\r\n\r\n**PQ is a diameter of a circle with the centre X and M is a point on the circumference of the circle, prove that $\\angle PMQ = 90^\\circ$ by vector method.**\r\n\r\n**Given:**\r\n* PMQ is a semi-circle with center X.\r\n* PQ is the diameter.\r\n\r\n\r\n\r\n**To Prove:** $\\angle PMQ = 90^\\circ$\r\n\r\n**Construction:** Join X to M.\r\n\r\n**Proof:**\r\n\r\n| Statement | Reason |\r\n| :--- | :--- |\r\n| i) $\\overrightarrow{QM} = \\overrightarrow{QX} + \\overrightarrow{XM} = \\overrightarrow{XM} - \\overrightarrow{XQ}$ | By Triangle Law of Vector Addition. |\r\n| ii) $\\overrightarrow{PM} = \\overrightarrow{PX} + \\overrightarrow{XM} = \\overrightarrow{XM} + \\overrightarrow{PX}$ | By Triangle Law of Vector Addition. |\r\n| iii) $\\overrightarrow{PM} \\cdot \\overrightarrow{QM} = (\\overrightarrow{XM} + \\overrightarrow{PX}) \\cdot (\\overrightarrow{XM} - \\overrightarrow{XQ})$ | Taking the dot product of (i) and (ii). |\r\n| iv) $\\overrightarrow{PM} \\cdot \\overrightarrow{QM} = (\\overrightarrow{XM} + \\overrightarrow{PX}) \\cdot (\\overrightarrow{XM} - \\overrightarrow{PX})$ | Radii are equal, so $|\\overrightarrow{XQ}| = |\\overrightarrow{PX}|$. Since they are in opposite directions, $\\overrightarrow{XQ} = -\\overrightarrow{PX}$. |\r\n| v) $\\overrightarrow{PM} \\cdot \\overrightarrow{QM} = |\\overrightarrow{XM}|^2 - |\\overrightarrow{PX}|^2$ | Using the formula $(a+b)(a-b) = a^2-b^2$. |\r\n| vi) $\\overrightarrow{PM} \\cdot \\overrightarrow{QM} = |\\overrightarrow{XM}|^2 - |\\overrightarrow{XM}|^2 = 0$ | $|\\overrightarrow{PX}| = |\\overrightarrow{XM}|$ (radii of the same circle). |\r\n| vii) $PM \\perp QM$, so $\\angle PMQ = 90^\\circ$ | The scalar product of the vectors is zero. |\r\n\r\n**Hence, the angle in a semi-circle is a right angle.**\r\n\r\n***\r\n\r\n<div id=\"isosceles\"></div>\r\n### 3. Median of an Isosceles Triangle\r\n\r\n**Prove by vector method that the median from the vertex of an isosceles triangle is perpendicular to the base.**\r\n\r\n**Given:**\r\n* In $\\triangle PQR$, $PQ = PR$.\r\n* S is the mid-point of the base QR.\r\n\r\n\r\n\r\n**To Prove:** $PS \\perp QR$\r\n\r\n**Proof:**\r\n\r\n| Statement | Reason |\r\n| :--- | :--- |\r\n| i) $\\overrightarrow{PS} = \\frac{1}{2}(\\overrightarrow{PQ} + \\overrightarrow{PR})$ | S is the mid-point of QR, so $\\overrightarrow{PS}$ is the median. |\r\n| ii) $\\overrightarrow{QR} = \\overrightarrow{PR} - \\overrightarrow{PQ}$ | By Triangle Law of Vector Addition ($\\overrightarrow{PQ} + \\overrightarrow{QR} = \\overrightarrow{PR}$). |\r\n| iii) $\\overrightarrow{PS} \\cdot \\overrightarrow{QR} = \\frac{1}{2}(\\overrightarrow{PQ} + \\overrightarrow{PR}) \\cdot (\\overrightarrow{PR} - \\overrightarrow{PQ})$ | Taking the dot product of statements (i) and (ii). |\r\n| iv) $\\overrightarrow{PS} \\cdot \\overrightarrow{QR} = \\frac{1}{2}(|\\overrightarrow{PR}|^2 - |\\overrightarrow{PQ}|^2)$ | Using the formula $(a+b)(a-b) = a^2-b^2$. |\r\n| v) $\\overrightarrow{PS} \\cdot \\overrightarrow{QR} = \\frac{1}{2}(|\\overrightarrow{PQ}|^2 - |\\overrightarrow{PQ}|^2) = 0$ | Given $PQ=PR$, so $|\\overrightarrow{PQ}| = |\\overrightarrow{PR}|$. |\r\n| vi) $PS \\perp QR$ | Since the scalar product is zero. |\r\n\r\n**Hence, the median to the base of an isosceles triangle is perpendicular to the base.**\r\n\r\n***\r\n\r\n<div id=\"rightAngled\"></div>\r\n### 4. Median to the Hypotenuse of a Right-Angled Triangle\r\n\r\n**In a right-angled triangle, prove by vector method that the length of the median to the hypotenuse is half the length of the hypotenuse.**\r\n\r\n**Given:**\r\n* In right-angled triangle PQR, $\\angle RPQ = 90^\\circ$.\r\n* S is the mid-point of the hypotenuse RQ.\r\n\r\n\r\n\r\n**To Prove:** $PS = QS = RS$ (which implies $PS = \\frac{1}{2}RQ$)\r\n\r\n**Proof:**\r\n\r\n| Statement | Reason |\r\n| :--- | :--- |\r\n| i) $\\angle RPQ = 90^\\circ$ | Given. |\r\n| ii) $\\overrightarrow{PR} \\cdot \\overrightarrow{PQ} = 0$ | The dot product of perpendicular vectors is zero. |\r\n| iii) $\\overrightarrow{PS} = \\frac{1}{2}(\\overrightarrow{PR} + \\overrightarrow{PQ})$ | By the mid-point theorem, as S is the midpoint of RQ. |\r\n| iv) $(\\overrightarrow{PS})^2 = \\frac{1}{4}(\\overrightarrow{PR} + \\overrightarrow{PQ})^2$ | Squaring both sides of statement (iii). |\r\n| v) $PS^2 = \\frac{1}{4}(|\\overrightarrow{PR}|^2 + 2\\overrightarrow{PR}\\cdot\\overrightarrow{PQ} + |\\overrightarrow{PQ}|^2)$ | Expanding the square. |\r\n| vi) $PS^2 = \\frac{1}{4}(PR^2 + 0 + PQ^2)$ | From statement (ii), $\\overrightarrow{PR} \\cdot \\overrightarrow{PQ} = 0$. |\r\n| vii) $PS^2 = \\frac{1}{4}(RQ^2)$ | In $\\triangle PQR$, by Pythagorean theorem, $PR^2+PQ^2=RQ^2$. |\r\n| viii) $PS = \\frac{1}{2}RQ$ | Taking the square root of both sides. |\r\n| ix) $PS = QS = RS$ | S is the mid-point of RQ, so $QS=RS=\\frac{1}{2}RQ$. |\r\n\r\n**Hence, the median to the hypotenuse is half the length of the hypotenuse.**\r\n\r\n***\r\n\r\n<div id=\"rhombus-perp\"></div>\r\n### 5. Diagonals of a Rhombus\r\n\r\n**Prove by vector method that the diagonals of a rhombus are perpendicular.**\r\n\r\n**Given:**\r\n* PQRS is a rhombus where $|\\overrightarrow{PQ}| = |\\overrightarrow{QR}| = |\\overrightarrow{RS}| = |\\overrightarrow{SP}|$.\r\n\r\n\r\n\r\n**To Prove:** $PR \\perp QS$\r\n\r\n**Proof:**\r\n\r\n| Statement | Reason |\r\n| :--- | :--- |\r\n| i) $\\overrightarrow{PR} = \\overrightarrow{PQ} + \\overrightarrow{PS}$ | By Parallelogram Law of Vector Addition. |\r\n| ii) $\\overrightarrow{QS} = \\overrightarrow{PS} - \\overrightarrow{PQ}$ | By Triangle Law of Vector Addition. |\r\n| iii) $\\overrightarrow{PR} \\cdot \\overrightarrow{QS} = (\\overrightarrow{PS} + \\overrightarrow{PQ}) \\cdot (\\overrightarrow{PS} - \\overrightarrow{PQ})$ | Taking the dot product of (i) and (ii). |\r\n| iv) $\\overrightarrow{PR} \\cdot \\overrightarrow{QS} = |\\overrightarrow{PS}|^2 - |\\overrightarrow{PQ}|^2$ | Using the formula $(a+b)(a-b) = a^2-b^2$. |\r\n| v) $\\overrightarrow{PR} \\cdot \\overrightarrow{QS} = |\\overrightarrow{PQ}|^2 - |\\overrightarrow{PQ}|^2 = 0$ | All sides of a rhombus are equal, so $|\\overrightarrow{PS}| = |\\overrightarrow{PQ}|$. |\r\n| vi) $PR \\perp QS$ | Since the scalar product is zero. |\r\n\r\n**Hence, the diagonals of a rhombus are perpendicular to each other.**\r\n\r\n***\r\n\r\n<div id=\'rectangle\'></div>\r\n### 6. Diagonals of a Rectangle\r\n\r\n**Prove by vector method that the diagonals of a rectangle are equal.**\r\n\r\n**Given:**\r\n* PQRS is a rectangle.\r\n* PR and QS are its diagonals.\r\n\r\n\r\n\r\n**To Prove:** $PR = QS$\r\n\r\n**Proof:**\r\n\r\n| Statement | Reason |\r\n| :--- | :--- |\r\n| i) $\\overrightarrow{PR} = \\overrightarrow{PQ} + \\overrightarrow{QR}$ | By Triangle Law of Vector Addition. |\r\n| ii) $PR^2 = (\\overrightarrow{PQ} + \\overrightarrow{QR})^2 = |\\overrightarrow{PQ}|^2 + 2\\overrightarrow{PQ}\\cdot\\overrightarrow{QR} + |\\overrightarrow{QR}|^2$ | Squaring both sides of (i). |\r\n| iii) $PR^2 = PQ^2 + QR^2$ | Since adjacent sides of a rectangle are perpendicular, $\\overrightarrow{PQ} \\cdot \\overrightarrow{QR} = 0$. |\r\n| iv) $\\overrightarrow{QS} = \\overrightarrow{QR} + \\overrightarrow{RS} = \\overrightarrow{QR} - \\overrightarrow{PQ}$ | By Triangle Law and since $\\overrightarrow{RS} = -\\overrightarrow{PQ}$. |\r\n| v) $QS^2 = (\\overrightarrow{QR} - \\overrightarrow{PQ})^2 = |\\overrightarrow{QR}|^2 - 2\\overrightarrow{QR}\\cdot\\overrightarrow{PQ} + |\\overrightarrow{PQ}|^2$ | Squaring both sides of (iv). |\r\n| vi) $QS^2 = QR^2 + PQ^2$ | Since $\\overrightarrow{QR} \\cdot \\overrightarrow{PQ} = 0$. |\r\n| vii) $PR^2 = QS^2$ | From statements (iii) and (vi). |\r\n| viii) $PR = QS$ | Taking the square root. |\r\n\r\n**Hence, the diagonals of a rectangle are equal.**\r\n\r\n***\r\n\r\n### 7. Diagonals of a Parallelogram\r\n\r\n**Prove by vector method that the diagonals of a parallelogram bisect each other.**\r\n\r\n**Given:**\r\n* OABC is a parallelogram.\r\n* M and N are the midpoints of diagonals AC and OB respectively.\r\n\r\n\r\n\r\n**To Prove:** The diagonals bisect each other (i.e., M and N are the same point, $\\overrightarrow{OM} = \\overrightarrow{ON}$).\r\n\r\n**Proof:**\r\n\r\n| Statement | Reason |\r\n| :--- | :--- |\r\n| i) $\\overrightarrow{OM} = \\frac{1}{2}(\\overrightarrow{OA} + \\overrightarrow{OC})$ | M is the midpoint of diagonal AC. |\r\n| ii) $\\overrightarrow{ON} = \\frac{1}{2}(\\overrightarrow{OB})$ | N is the midpoint of diagonal OB. |\r\n| iii) $\\overrightarrow{OB} = \\overrightarrow{OA} + \\overrightarrow{OC}$ | By Parallelogram Law of Vector Addition. |\r\n| iv) $\\overrightarrow{ON} = \\frac{1}{2}(\\overrightarrow{OA} + \\overrightarrow{OC})$ | Substituting statement (iii) into (ii). |\r\n| v) $\\overrightarrow{OM} = \\overrightarrow{ON}$ | From statements (i) and (iv). |\r\n\r\n**Since the position vectors of the mid-points of both diagonals are the same, the mid-points coincide. Hence, the diagonals of a parallelogram bisect each other.**\r\n\r\n***\r\n\r\n### 8. Mid-Point Theorem for Quadrilaterals\r\n\r\n**Prove by vector method that the quadrilateral formed by joining the mid-points of the sides of any quadrilateral is a parallelogram.**\r\n\r\n**Given:**\r\n* ABCD is a quadrilateral.\r\n* P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA respectively.\r\n\r\n\r\n\r\n**To Prove:** PQRS is a parallelogram.\r\n\r\n**Construction:** Join diagonal AC.\r\n\r\n**Proof:**\r\n\r\n| Statement | Reason |\r\n| :--- | :--- |\r\n| i) In $\\triangle ABC$, $\\overrightarrow{PQ} = \\frac{1}{2}\\overrightarrow{AC}$ | The line segment joining the mid-points of two sides of a triangle is parallel to and half the length of the third side. |\r\n| ii) In $\\triangle ADC$, $\\overrightarrow{SR} = \\frac{1}{2}\\overrightarrow{AC}$ | Same reason as above. |\r\n| iii) $\\overrightarrow{PQ} = \\overrightarrow{SR}$ | From statements (i) and (ii). This implies $PQ \\parallel SR$ and $PQ = SR$. |\r\n| iv) PQRS is a parallelogram. | A quadrilateral is a parallelogram if one pair of opposite sides is equal and parallel. |\r\n\r\n**Hence, the quadrilateral formed by joining the mid-points is a parallelogram.**\r\n\r\n***\r\n\r\n### 9. Section Formula Proofs\r\n\r\n#### a) Internal Division\r\n\r\n**If the position vectors of points P and Q are $\\overrightarrow{p}$ and $\\overrightarrow{q}$ respectively, and point M divides the line segment PQ internally in the ratio $m_1:m_2$, prove that the position vector of M is $\\overrightarrow{OM} = \\frac{m_1\\overrightarrow{q} + m_2\\overrightarrow{p}}{m_1 + m_2}$.**\r\n\r\n\r\n\r\n**Proof:**\r\n\r\n| Statement | Reason |\r\n| :--- | :--- |\r\n| i) $\\frac{PM}{MQ} = \\frac{m_1}{m_2}$ | Given ratio. |\r\n| ii) $m_2 \\overrightarrow{PM} = m_1 \\overrightarrow{MQ}$ | Cross-multiplication. Since P, M, Q are collinear and M is internal, the vectors are in the same direction. |\r\n| iii) $m_2(\\overrightarrow{OM} - \\overrightarrow{OP}) = m_1(\\overrightarrow{OQ} - \\overrightarrow{OM})$ | By Triangle Law of Vector Addition. |\r\n| iv) $m_2\\overrightarrow{OM} - m_2\\overrightarrow{p} = m_1\\overrightarrow{q} - m_1\\overrightarrow{OM}$ | Substituting given position vectors. |\r\n| v) $m_1\\overrightarrow{OM} + m_2\\overrightarrow{OM} = m_1\\overrightarrow{q} + m_2\\overrightarrow{p}$ | Rearranging terms. |\r\n| vi) $\\overrightarrow{OM}(m_1 + m_2) = m_1\\overrightarrow{q} + m_2\\overrightarrow{p}$ | Taking $\\overrightarrow{OM}$ common. |\r\n| vii) $\\overrightarrow{OM} = \\frac{m_1\\overrightarrow{q} + m_2\\overrightarrow{p}}{m_1 + m_2}$ | Dividing by $(m_1+m_2)$. |\r\n\r\n#### b) External Division\r\n\r\n**If point M divides the line segment AB externally in the ratio $m:n$, prove that its position vector is $\\overrightarrow{OM} = \\frac{m\\overrightarrow{b} - n\\overrightarrow{a}}{m - n}$.**\r\n\r\n\r\n\r\n**Proof:**\r\n\r\n| Statement | Reason |\r\n| :--- | :--- |\r\n| i) $\\frac{AM}{BM} = \\frac{m}{n}$ | Given ratio. |\r\n| ii) $n \\overrightarrow{AM} = m \\overrightarrow{BM}$ | Cross-multiplication. Vectors are in the same direction. |\r\n| iii) $n(\\overrightarrow{OM} - \\overrightarrow{OA}) = m(\\overrightarrow{OM} - \\overrightarrow{OB})$ | By Triangle Law of Vector Addition. |\r\n| iv) $n\\overrightarrow{OM} - n\\overrightarrow{a} = m\\overrightarrow{OM} - m\\overrightarrow{b}$ | Substituting position vectors $\\overrightarrow{a}$ and $\\overrightarrow{b}$. |\r\n| v) $m\\overrightarrow{b} - n\\overrightarrow{a} = m\\overrightarrow{OM} - n\\overrightarrow{OM}$ | Rearranging terms. |\r\n| vi) $m\\overrightarrow{b} - n\\overrightarrow{a} = \\overrightarrow{OM}(m - n)$ | Taking $\\overrightarrow{OM}$ common. |\r\n| vii) $\\overrightarrow{OM} = \\frac{m\\overrightarrow{b} - n\\overrightarrow{a}}{m - n}$ | Dividing by $(m-n)$. |\r\n\r\n---\r\n\r\n## 🔢 Numerical Problems & Solutions\r\n\r\nThis section contains worked-out solutions for numerical problems applying the concepts of position vectors and section formulas.\r\n\r\n***\r\n\r\n### Problem 1: Finding the Centroid (from Median)\r\n**In $\\triangle ABC$, D is the midpoint of AC. If the position vectors of B and D are $3\\overrightarrow{i} + 7\\overrightarrow{j}$ and $3\\overrightarrow{i} - 2\\overrightarrow{j}$ respectively, find the position vector of the centroid G.**\r\n\r\n\r\n\r\n**Solution:**\r\nLet O be the origin.\r\nGiven: $\\overrightarrow{OB} = 3\\overrightarrow{i} + 7\\overrightarrow{j}$ and $\\overrightarrow{OD} = 3\\overrightarrow{i} - 2\\overrightarrow{j}$.\r\nSince D is the mid-point of AC: $\\overrightarrow{OD} = \\frac{\\overrightarrow{OA} + \\overrightarrow{OC}}{2} \\implies \\overrightarrow{OA} + \\overrightarrow{OC} = 2\\overrightarrow{OD} = 6\\overrightarrow{i} - 4\\overrightarrow{j}$.\r\nThe position vector of the centroid G is:\r\n$$\r\n\\begin{aligned}\r\n\\overrightarrow{OG} &= \\frac{\\overrightarrow{OA} + \\overrightarrow{OB} + \\overrightarrow{OC}}{3} = \\frac{(\\overrightarrow{OA} + \\overrightarrow{OC}) + \\overrightarrow{OB}}{3} \\\\\r\n&= \\frac{(6\\overrightarrow{i} - 4\\overrightarrow{j}) + (3\\overrightarrow{i} + 7\\overrightarrow{j})}{3} \\\\\r\n&= \\frac{9\\overrightarrow{i} + 3\\overrightarrow{j}}{3} = 3\\overrightarrow{i} + \\overrightarrow{j}\r\n\\end{aligned}\r\n$$\r\n**The position vector of G is $3\\overrightarrow{i} + \\overrightarrow{j}$.**\r\n\r\n***\r\n\r\n### Problem 2: Finding the Centroid (from Vertices)\r\n**If the position vectors of the vertices A, B, and C of $\\triangle ABC$ are $(3\\overrightarrow{i} + 5\\overrightarrow{j})$, $(5\\overrightarrow{i} - \\overrightarrow{j})$ and $(\\overrightarrow{i} - 8\\overrightarrow{j})$ respectively, find the position vector of the centroid G.**\r\n\r\n**Solution:**\r\nGiven vertices:\r\n* $\\overrightarrow{OA} = 3\\overrightarrow{i} + 5\\overrightarrow{j}$\r\n* $\\overrightarrow{OB} = 5\\overrightarrow{i} - \\overrightarrow{j}$\r\n* $\\overrightarrow{OC} = \\overrightarrow{i} - 8\\overrightarrow{j}$\r\n\r\nUsing the centroid formula:\r\n$$\r\n\\begin{aligned}\r\n\\overrightarrow{OG} &= \\frac{\\overrightarrow{OA} + \\overrightarrow{OB} + \\overrightarrow{OC}}{3} \\\\\r\n&= \\frac{(3\\overrightarrow{i} + 5\\overrightarrow{j}) + (5\\overrightarrow{i} - \\overrightarrow{j}) + (\\overrightarrow{i} - 8\\overrightarrow{j})}{3} \\\\\r\n&= \\frac{(3+5+1)\\overrightarrow{i} + (5-1-8)\\overrightarrow{j}}{3} \\\\\r\n&= \\frac{9\\overrightarrow{i} - 4\\overrightarrow{j}}{3} \\\\\r\n&= 3\\overrightarrow{i} - \\frac{4}{3}\\overrightarrow{j}\r\n\\end{aligned}\r\n$$\r\n**The position vector of G is $3\\overrightarrow{i} - \\frac{4}{3}\\overrightarrow{j}$.**\r\n\r\n***\r\n\r\n### Problem 3: Finding a Mid-point Vector\r\n**If the position vectors of two points A and B are $\\begin{pmatrix} 3 \\\\ 4 \\end{pmatrix}$ and $\\begin{pmatrix} 4 \\\\ 5 \\end{pmatrix}$ respectively, find the position vector of K, the mid-point of AB.**\r\n\r\n**Solution:**\r\nGiven: $\\overrightarrow{OA} = \\begin{pmatrix} 3 \\\\ 4 \\end{pmatrix}$ and $\\overrightarrow{OB} = \\begin{pmatrix} 4 \\\\ 5 \\end{pmatrix}$.\r\n$$\r\n\\begin{aligned}\r\n\\overrightarrow{OK} &= \\frac{\\overrightarrow{OA} + \\overrightarrow{OB}}{2} \\\\\r\n&= \\frac{1}{2} \\left[ \\begin{pmatrix} 3 \\\\ 4 \\end{pmatrix} + \\begin{pmatrix} 4 \\\\ 5 \\end{pmatrix} \\right] = \\frac{1}{2} \\begin{pmatrix} 7 \\\\ 9 \\end{pmatrix} = \\begin{pmatrix} 3.5 \\\\ 4.5 \\end{pmatrix}\r\n\\end{aligned}\r\n$$\r\n**The position vector of K is $\\begin{pmatrix} 3.5 \\\\ 4.5 \\end{pmatrix}$.**\r\n\r\n***\r\n\r\n### Problem 4: Finding the Position Vector of P where MP = PN\r\n**The position vectors of M and N are $9\\overrightarrow{i} + 3\\overrightarrow{j}$ and $3\\overrightarrow{i} + 5\\overrightarrow{j}$ respectively. Find the position vector of a point P such that $\\overrightarrow{MP} = \\overrightarrow{PN}$.**\r\n\r\n**Solution:**\r\nThe condition $\\overrightarrow{MP} = \\overrightarrow{PN}$ means that P is the mid-point of the line segment MN.\r\nGiven:\r\n* $\\overrightarrow{OM} = 9\\overrightarrow{i} + 3\\overrightarrow{j}$\r\n* $\\overrightarrow{ON} = 3\\overrightarrow{i} + 5\\overrightarrow{j}$\r\n\r\nUsing the mid-point formula for P:\r\n$$\r\n\\begin{aligned}\r\n\\overrightarrow{OP} &= \\frac{\\overrightarrow{OM} + \\overrightarrow{ON}}{2} \\\\\r\n&= \\frac{(9\\overrightarrow{i} + 3\\overrightarrow{j}) + (3\\overrightarrow{i} + 5\\overrightarrow{j})}{2} \\\\\r\n&= \\frac{12\\overrightarrow{i} + 8\\overrightarrow{j}}{2} \\\\\r\n&= 6\\overrightarrow{i} + 4\\overrightarrow{j}\r\n\\end{aligned}\r\n$$\r\n**The position vector of P is $6\\overrightarrow{i} + 4\\overrightarrow{j}$.**\r\n\r\n***\r\n\r\n### Problem 5: Finding a Vector from a Collinear Relationship\r\n**In the given figure, $\\overrightarrow{OA} = \\overrightarrow{a}$ and $\\overrightarrow{OB} = \\overrightarrow{b}$. If $\\overrightarrow{AC} = 3\\overrightarrow{AB}$, find $\\overrightarrow{OC}$.**\r\n\r\n\r\n\r\n**Solution:**\r\nWe need to find $\\overrightarrow{OC}$. Using the triangle law:\r\n$$ \\overrightarrow{OC} = \\overrightarrow{OA} + \\overrightarrow{AC} $$\r\nWe are given $\\overrightarrow{AC} = 3\\overrightarrow{AB}$. First, let\'s find $\\overrightarrow{AB}$:\r\n$$ \\overrightarrow{AB} = \\overrightarrow{OB} - \\overrightarrow{OA} = \\overrightarrow{b} - \\overrightarrow{a} $$\r\nNow substitute this into the expression for $\\overrightarrow{AC}$:\r\n$$ \\overrightarrow{AC} = 3(\\overrightarrow{b} - \\overrightarrow{a}) $$\r\nFinally, find $\\overrightarrow{OC}$:\r\n$$\r\n\\begin{aligned}\r\n\\overrightarrow{OC} &= \\overrightarrow{OA} + \\overrightarrow{AC} \\\\\r\n&= \\overrightarrow{a} + 3(\\overrightarrow{b} - \\overrightarrow{a}) \\\\\r\n&= \\overrightarrow{a} + 3\\overrightarrow{b} - 3\\overrightarrow{a} \\\\\r\n&= 3\\overrightarrow{b} - 2\\overrightarrow{a}\r\n\\end{aligned}\r\n$$\r\n**The position vector $\\overrightarrow{OC}$ is $3\\overrightarrow{b} - 2\\overrightarrow{a}$.**', 1, '2025-07-25 01:11:38', '2025-09-18 09:23:05'),
(149, 14247, '# Grade 10 Optional Mathematics: Continuity\r\n## Unit: Calculus | Chapter: Continuity and Discontinuity | Complete Solutions\r\n\r\n---\r\n\r\n## 🔑 Key Concepts & Definitions\r\n\r\nBefore attempting the problems, it\'s important to understand the core concepts of limits and continuity.\r\n\r\n### 1. Limits\r\n\r\n* **Left-Hand Limit (LHL):** The value that a function `f(x)` approaches as `x` gets closer to a point `a` from the left side (values less than `a`). It is denoted as:\r\n    $$ \\lim_{x \\to a^-} f(x) $$\r\n\r\n* **Right-Hand Limit (RHL):** The value that a function `f(x)` approaches as `x` gets closer to a point `a` from the right side (values greater than `a`). It is denoted as:\r\n    $$ \\lim_{x \\to a^+} f(x) $$\r\n\r\n* **Existence of a Limit:** For the limit of a function `f(x)` to exist at `x = a`, the left-hand limit must equal the right-hand limit.\r\n    $$ \\lim_{x \\to a^-} f(x) = \\lim_{x \\to a^+} f(x) $$\r\n\r\n### 2. Functional Value\r\n\r\n* **Functional Value:** The value of the function `f(x)` at a specific point `x = a`. It is denoted as `f(a)`. This is the direct output of the function when `x` is exactly equal to `a`.\r\n\r\n### 3. Continuity and Discontinuity\r\n\r\n* **Continuity:** A function `f(x)` is said to be **continuous** at a point `x = a` if three conditions are met:\r\n    1.  The left-hand limit exists: $\\lim_{x \\to a^-} f(x)$\r\n    2.  The right-hand limit exists: $\\lim_{x \\to a^+} f(x)$\r\n    3.  Both limits are equal to the functional value: $\\lim_{x \\to a^-} f(x) = \\lim_{x \\to a^+} f(x) = f(a)$\r\n\r\n    Essentially, the graph of the function has no breaks, jumps, or holes at that point.\r\n\r\n* **Discontinuity:** A function is **discontinuous** at `x = a` if any of the three conditions for continuity are not met.\r\n\r\n### 4. Intervals\r\n\r\n* **Open Interval:** An interval that does not include its endpoints. It is denoted with parentheses, e.g., `(a, b)`.\r\n* **Closed Interval:** An interval that includes its endpoints. It is denoted with square brackets, e.g., `[a, b]`.\r\n\r\n---\r\n\r\n## 📚 Exercise Solutions\r\n\r\n### Problem 1: Investigating Continuity\r\n**Given the function f(x) = 3x + 2:**\r\n**a) Find f(2.001), f(2.0001), f(1.999), and f(1.9999).**\r\n**b) Find $\\lim_{x \\to 2^-} f(x)$, $\\lim_{x \\to 2^+} f(x)$, and f(2).**\r\n**c) Is the function continuous at x = 2?**\r\n\r\n**Solution:**\r\nHere, $f(x) = 3x + 2$.\r\n\r\n**a) Calculating values near x = 2:**\r\n* $f(2.001) = 3(2.001) + 2 = 6.003 + 2 = 8.003$\r\n* $f(2.0001) = 3(2.0001) + 2 = 6.0003 + 2 = 8.0003$\r\n* $f(1.999) = 3(1.999) + 2 = 5.997 + 2 = 7.997$\r\n* $f(1.9999) = 3(1.9999) + 2 = 5.9997 + 2 = 7.9997$\r\n    \r\n*As x approaches 2 from both sides, f(x) approaches 8.*\r\n\r\n**b) Finding the limits and functional value:**\r\n\r\n* **Left-Hand Limit (LHL):**\r\n    $$ \\lim_{x \\to 2^-} f(x) = \\lim_{x \\to 2^-} (3x + 2) = 3(2) + 2 = 8 $$\r\n* **Right-Hand Limit (RHL):**\r\n    $$ \\lim_{x \\to 2^+} f(x) = \\lim_{x \\to 2^+} (3x + 2) = 3(2) + 2 = 8 $$\r\n* **Functional Value:**\r\n    $$ f(2) = 3(2) + 2 = 8 $$\r\n\r\n**c) Checking for continuity:**\r\nSince LHL = RHL = f(2) = 8, all three conditions for continuity are met.\r\n**Yes, the function is continuous at x = 2.**\r\n\r\n***\r\n\r\n### Problem 2: Continuity of a Rational Function\r\n**Given the function $f(x) = \\frac{x^2 - 9}{x - 3}$:**\r\n**a) Find f(2.999) and f(3.001).**\r\n**b) Are f(2.999) and f(3.001) approximately equal?**\r\n**c) Discuss the continuity of f(x) at x = 3.**\r\n\r\n**Solution:**\r\nHere, $f(x) = \\frac{x^2 - 9}{x - 3}$.\r\n\r\n**a) Calculating values near x = 3:**\r\n* $f(2.999) = \\frac{(2.999)^2 - 9}{2.999 - 3} = \\frac{8.994001 - 9}{-0.001} = \\frac{-0.005999}{-0.001} = 5.999$\r\n* $f(3.001) = \\frac{(3.001)^2 - 9}{3.001 - 3} = \\frac{9.006001 - 9}{0.001} = \\frac{0.006001}{0.001} = 6.001$\r\n\r\n**b) Approximation:**\r\nYes, after approximation, both $f(2.999) \\approx 6$ and $f(3.001) \\approx 6$.\r\n\r\n**c) Discussing continuity at x = 3:**\r\n* **Limiting Value:**\r\n    $$ \\lim_{x \\to 3} f(x) = \\lim_{x \\to 3} \\frac{x^2 - 9}{x - 3} \\quad (\\text{form } \\frac{0}{0}) $$\r\n    $$ = \\lim_{x \\to 3} \\frac{(x-3)(x+3)}{x-3} = \\lim_{x \\to 3} (x+3) = 3+3=6 $$\r\n* **Functional Value:**\r\n    The function $f(3) = \\frac{3^2 - 9}{3 - 3} = \\frac{0}{0}$, which is undefined.\r\n\r\n**Conclusion:** Since the functional value $f(3)$ is undefined, the function is **discontinuous** at $x = 3$.\r\n\r\n***\r\n\r\n### Problem 3: Examining Continuity of Piecewise Functions\r\n\r\n**Examine the continuity or discontinuity of the following functions at the specified points.**\r\n\r\n**a) $f(x) = \\begin{cases} 2 - x^2, & \\text{if } x \\leq 2 \\\\ x - 4, & \\text{if } x > 2 \\end{cases}$ at $x = 2$**\r\n\r\n**Solution:**\r\n* **LHL:** $\\lim_{x \\to 2^-} f(x) = \\lim_{x \\to 2^-} (2 - x^2) = 2 - 2^2 = -2$.\r\n* **RHL:** $\\lim_{x \\to 2^+} f(x) = \\lim_{x \\to 2^+} (x - 4) = 2 - 4 = -2$.\r\n* **Functional Value:** $f(2) = 2 - 2^2 = -2$.\r\n\r\nSince LHL = RHL = f(2) = -2, the function is **continuous** at $x = 2$.\r\n\r\n**b) $f(x) = \\begin{cases} x, & \\text{if } x < 0 \\\\ 0, & \\text{if } x = 0 \\\\ x^2, & \\text{if } x > 0 \\end{cases}$ at $x = 0$**\r\n\r\n**Solution:**\r\n* **LHL:** $\\lim_{x \\to 0^-} f(x) = \\lim_{x \\to 0^-} (x) = 0$.\r\n* **RHL:** $\\lim_{x \\to 0^+} f(x) = \\lim_{x \\to 0^+} (x^2) = 0^2 = 0$.\r\n* **Functional Value:** $f(0) = 0$.\r\n\r\nSince LHL = RHL = f(0) = 0, the function is **continuous** at $x = 0$.\r\n\r\n**c) $f(x) = \\begin{cases} \\frac{2}{5 - x}, & \\text{if } x \\leq 3 \\\\ 5 - x, & \\text{if } x > 3 \\end{cases}$ at $x = 3$**\r\n\r\n**Solution:**\r\n* **LHL:** $\\lim_{x \\to 3^-} f(x) = \\lim_{x \\to 3^-} \\frac{2}{5 - x} = \\frac{2}{5 - 3} = 1$.\r\n* **RHL:** $\\lim_{x \\to 3^+} f(x) = \\lim_{x \\to 3^+} (5 - x) = 5 - 3 = 2$.\r\n\r\nSince LHL $\\neq$ RHL, the function is **discontinuous** at $x = 3$.\r\n\r\n**d) $f(x) = \\begin{cases} x^2 + 2, & \\text{if } x < 5 \\\\ 6x - 3, & \\text{if } x = 5 \\\\ 3x + 12, & \\text{if } x > 5 \\end{cases}$ at $x = 5$**\r\n\r\n**Solution:**\r\n* **LHL:** $\\lim_{x \\to 5^-} f(x) = \\lim_{x \\to 5^-} (x^2 + 2) = 5^2 + 2 = 27$.\r\n* **RHL:** $\\lim_{x \\to 5^+} f(x) = \\lim_{x \\to 5^+} (3x + 12) = 3(5) + 12 = 27$.\r\n* **Functional Value:** $f(5) = 6(5) - 3 = 30 - 3 = 27$.\r\n\r\nSince LHL = RHL = f(5) = 27, the function is **continuous** at $x = 5$.\r\n\r\n***\r\n\r\n### Problem 4: Finding an Unknown Value for Continuity\r\n**A function is defined as $f(x) = \\begin{cases} kx + 3, & \\text{if } x \\ge 2 \\\\ 3x - 1, & \\text{if } x < 2 \\end{cases}$. Find the value of k so that f(x) is continuous at x = 2.**\r\n\r\n**Solution:**\r\nFor the function to be continuous at $x=2$, we must have LHL = RHL = f(2).\r\n\r\n* **LHL:** (approaching 2 from the left, so $x < 2$)\r\n    $$ \\lim_{x \\to 2^-} f(x) = \\lim_{x \\to 2^-} (3x - 1) = 3(2) - 1 = 5 $$\r\n* **RHL:** (approaching 2 from the right, so $x > 2$)\r\n    $$ \\lim_{x \\to 2^+} f(x) = \\lim_{x \\to 2^+} (kx + 3) = k(2) + 3 = 2k + 3 $$\r\n* **Functional Value:** (at $x=2$)\r\n    $$ f(2) = k(2) + 3 = 2k + 3 $$\r\n\r\nNow, set LHL equal to RHL:\r\n$$\r\n\\begin{aligned}\r\n5 &= 2k + 3 \\\\\r\n2 &= 2k \\\\\r\nk &= 1\r\n\\end{aligned}\r\n$$\r\n**The value of k must be 1 for the function to be continuous at x = 2.**\r\n\r\n***\r\n\r\n### Problem 5: Finding an Unknown Value for Continuity (Rational Function)\r\n**A function is defined as $f(x) = \\begin{cases} \\frac{2x^2 - 18}{x - 3}, & \\text{for } x \\neq 3 \\\\ k, & \\text{for } x = 3 \\end{cases}$. Find the value of k so that f(x) is continuous at x = 3.**\r\n\r\n**Solution:**\r\nFor the function to be continuous at $x=3$, the limit of the function as $x \\to 3$ must be equal to the functional value $f(3)$.\r\n\r\n* **Limiting Value:**\r\n    $$ \\lim_{x \\to 3} f(x) = \\lim_{x \\to 3} \\frac{2x^2 - 18}{x - 3} $$\r\n    This is in the indeterminate form $\\frac{0}{0}$. We simplify the expression:\r\n    $$ = \\lim_{x \\to 3} \\frac{2(x^2 - 9)}{x - 3} = \\lim_{x \\to 3} \\frac{2(x-3)(x+3)}{x-3} = \\lim_{x \\to 3} 2(x+3) = 2(3+3) = 12 $$\r\n* **Functional Value:**\r\n    $$ f(3) = k $$\r\n\r\nTo ensure continuity, we set the limiting value equal to the functional value:\r\n$$ f(3) = \\lim_{x \\to 3} f(x) $$\r\n$$ k = 12 $$\r\n**The value of k must be 12 for the function to be continuous at x = 3.**\r\n\r\n***\r\n\r\n### Problem 6: Redefining a Function to be Continuous\r\n**A function is defined as $f(x) = \\begin{cases} 2x - 3, & \\text{if } x < 2 \\\\ 2, & \\text{if } x = 2 \\\\ 3x - 5, & \\text{if } x > 2 \\end{cases}$. Is f(x) continuous at x = 2? If not, how can it be made continuous?**\r\n\r\n**Solution:**\r\nLet\'s check the three conditions for continuity at $x=2$.\r\n\r\n* **LHL:** $\\lim_{x \\to 2^-} f(x) = \\lim_{x \\to 2^-} (2x - 3) = 2(2) - 3 = 1$.\r\n* **RHL:** $\\lim_{x \\to 2^+} f(x) = \\lim_{x \\to 2^+} (3x - 5) = 3(2) - 5 = 1$.\r\n* **Functional Value:** $f(2) = 2$.\r\n\r\nHere, LHL = RHL = 1, so the limit exists and is equal to 1. However, the functional value $f(2) = 2$.\r\nSince $\\lim_{x \\to 2} f(x) \\neq f(2)$, the function is **discontinuous** at $x=2$.\r\n\r\nTo make the function continuous, the functional value must be equal to the limit. We must redefine $f(2)$ to be 1.\r\n\r\nThe redefined continuous function is:\r\n$$ f(x) = \\begin{cases} 2x - 3, & \\text{if } x < 2 \\\\ 1, & \\text{if } x = 2 \\\\ 3x - 5, & \\text{if } x > 2 \\end{cases} $$', 1, '2025-07-25 01:11:38', '2025-09-18 09:23:05'),
(150, 14242, '\n# Grade 10 Compulsory Mathematics: Statistics Study Guide\n## Unit: Statistics | Chapter: Measures of Dispersion | Complete Solutions\n\n---\n\n## 🎯 Chapter Introduction: Understanding Dispersion\n\nThis guide covers the fundamental concepts of **Measures of Dispersion** in statistics. Dispersion helps us understand how spread out or scattered a set of data is. Think of it as measuring the \"variety\" or \"variability\" in your data. A low dispersion means the data points are clustered together, while high dispersion indicates they are spread far apart. 📊\n\n### Key Learning Areas\n* **Defining Dispersion:** What it means for data to be scattered.\n* **Absolute vs. Relative Measures:** Understanding the two main types of dispersion measures.\n* **Quartile Deviation (QD):** A measure of dispersion based on the middle 50% of the data.\n* **Coefficient of Quartile Deviation:** A relative measure for comparing dispersion between different datasets.\n* **Calculations for Continuous Data:** Step-by-step methods to calculate QD and its coefficient for frequency distributions.\n\n---\n\n## 🔑 Key Concepts & Definitions\n\nBefore jumping into calculations, let\'s clarify the core ideas.\n\n### 1. What is Dispersion?\n**Dispersion** simply means the extent to which numerical data is likely to vary or scatter around an average value. In other words, it\'s the measure of the **variability** or **fluctuation** within a dataset.\n\n---\n\n### 2. Types of Dispersion Measures\nThere are two main categories for measuring dispersion:\n\n* **Absolute Measure of Dispersion:** These measures express the variability in the **same units** as the original data (e.g., kg, marks, inches). They are used for describing the amount of spread in a single dataset.\n    * Examples: Range, **Quartile Deviation**, Mean Deviation, Standard Deviation.\n\n* **Relative Measure of Dispersion:** These are unit-less ratios used for **comparing the variability** of two or more datasets, especially when their units or averages are different.\n    * Examples: Coefficient of Range, **Coefficient of Quartile Deviation**, Coefficient of Variation.\n\n---\n\n### 3. Quartile Deviation (QD)\n**Quartile Deviation (QD)**, also known as the **Semi-Interquartile Range**, measures the average spread of the middle 50% of the data around the median. It is calculated as half the difference between the upper quartile ($Q_3$) and the lower quartile ($Q_1$).\n\n* **Formula:**\n    $$\n    \\text{Quartile Deviation (QD)} = \\frac{Q_3 - Q_1}{2}\n    $$\n\n---\n\n### 4. Coefficient of Quartile Deviation\nThe **Coefficient of Quartile Deviation** is a relative measure. It expresses the quartile deviation as a percentage of the average of the upper and lower quartiles, making it perfect for comparing the consistency of different datasets.\n\n* **Formula:**\n    $$\n    \\text{Coefficient of QD} = \\frac{Q_3 - Q_1}{Q_3 + Q_1}\n    $$\n\n---\n\n## 📚 Exercise Solutions\n\n### Short Questions\n\n***\n\n#### Question 1: What is dispersion?\n**Answer:** Dispersion refers to the scattering, variability, deviation, or fluctuation of data points around a central value.\n\n***\n\n#### Question 2: Write the various measures of dispersion.\n**Answer:**\nThe various measures of dispersion are:\n1.  **Absolute Measures:** Range, Quartile Deviation, Mean Deviation, and Standard Deviation.\n2.  **Relative Measures:** Coefficient of Range, Coefficient of Quartile Deviation, Coefficient of Mean Deviation, Coefficient of Variation.\n\n***\n\n#### Question 3: Define quartile deviation and write its formula.\n**Answer:**\nHalf of the difference between the upper quartile ($Q_3$) and the lower quartile ($Q_1$) is called **quartile deviation (QD)** or the semi-interquartile range.\nIts formula is:\n$$\n\\text{Quartile Deviation (QD)} = \\frac{Q_3 - Q_1}{2}\n$$\n\n***\n\n#### Question 4: What do you mean by the coefficient of quartile deviation?\n**Answer:**\nThe coefficient of quartile deviation is the relative measure of dispersion based on the lower and upper quartiles. It is used for comparison purposes.\nIts formula is:\n$$\n\\text{Coefficient of Quartile Deviation} = \\frac{Q_3 - Q_1}{Q_3 + Q_1}\n$$\n\n***\n\n#### Question 5: Differentiate between quartile deviation and its coefficient.\n**Answer:**\nQuartile deviation is an **absolute measure** of dispersion, calculated as $\\frac{Q_3 - Q_1}{2}$, and is expressed in the same units as the data. In contrast, the coefficient of quartile deviation is a **relative measure**, calculated as $\\frac{Q_3 - Q_1}{Q_3 + Q_1}$, which is a unit-less ratio used to compare the variability of different datasets.\n\n***\n\n#### Question 6: In continuous data, if $Q_1 = 40$ and $Q_3 = 60$, find the quartile deviation.\n**Solution:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **Find Quartile Deviation (QD)** | $QD = \\frac{Q_3 - Q_1}{2} = \\frac{60 - 40}{2} = \\mathbf{10}$ |\n\n***\n\n#### Question 7: In a continuous series, the lower quartile is 25 and its quartile deviation is 10. Find the upper quartile.\n**Solution:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **Find Upper Quartile ($Q_3$)** | $10 = \\frac{Q_3 - 25}{2} \\implies 20 = Q_3 - 25 \\implies Q_3 = \\mathbf{45}$ |\n\n***\n\n#### Question 8: In a continuous series, the coefficient of quartile deviation is $\\frac{1}{2}$ and its upper quartile is 60. Find the first quartile.\n**Solution:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **Find First Quartile ($Q_1$)** | $\\frac{1}{2} = \\frac{60 - Q_1}{60 + Q_1} \\implies 60 + Q_1 = 120 - 2Q_1 \\implies 3Q_1 = 60 \\implies Q_1 = \\mathbf{20}$ |\n\n***\n\n#### Question 9: If the coefficient of quartile deviation is $\\frac{3}{7}$ and $Q_1 = 20$, find $Q_3$.\n**Solution:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **Find Upper Quartile ($Q_3$)** | $\\frac{3}{7} = \\frac{Q_3 - 20}{Q_3 + 20} \\implies 3Q_3 + 60 = 7Q_3 - 140 \\implies 200 = 4Q_3 \\implies Q_3 = \\mathbf{50}$ |\n\n---\n\n### Long Questions\n\n***\n\n#### Question 10: Find the quartile deviation and its coefficient from the following data.\n\n| Marks obtained | No. of students |\n|:---:|:---:|\n| 60-65 | 7 |\n| 65-70 | 5 |\n| 70-75 | 8 |\n| 75-80 | 4 |\n| 80-85 | 3 |\n| 85-90 | 3 |\n\n**Solution:**\n1.  **Cumulative Frequency Table:**\n\n| Marks obtained | Frequency (f) | Cumulative Frequency (c.f.) |\n|:---:|:---:|:---:|\n| 60-65 | 7 | 7 |\n| 65-70 | 5 | 12 |\n| 70-75 | 8 | 20 |\n| 75-80 | 4 | 24 |\n| 80-85 | 3 | 27 |\n| 85-90 | 3 | 30 |\n| **Total** | **N = 30** | |\n\n2.  **Find Quartiles:**\n    * **For $Q_1$**: Position is $(\\frac{30}{4})^{th} = 7.5^{th}$ item. $Q_1$ class is **65-70**.\n        $Q_1 = 65 + \\frac{7.5 - 7}{5} \\times 5 = \\mathbf{65.5}$\n    * **For $Q_3$**: Position is $(\\frac{3 \\times 30}{4})^{th} = 22.5^{th}$ item. $Q_3$ class is **75-80**.\n        $Q_3 = 75 + \\frac{22.5 - 20}{4} \\times 5 = \\mathbf{78.125}$\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. Quartile Deviation (QD)** | $QD = \\frac{78.125 - 65.5}{2} = \\mathbf{6.3125}$ |\n| **2. Coefficient of QD** | $\\text{Coefficient} = \\frac{78.125 - 65.5}{78.125 + 65.5} = \\frac{12.625}{143.625} \\approx \\mathbf{0.088}$ |\n\n***\n\n#### Question 11: Find the quartile deviation and its coefficient from the following data.\n\n| Weight (kg) | No. of person |\n|:---:|:---:|\n| 20-30 | 5 |\n| 30-40 | 15 |\n| 40-50 | 10 |\n| 50-60 | 8 |\n| 60-70 | 6 |\n| 70-80 | 2 |\n\n**Solution:**\n1.  **Cumulative Frequency Table:**\n\n| Weight (kg) | Frequency (f) | c.f. |\n|:---:|:---:|:---:|\n| 20-30 | 5 | 5 |\n| 30-40 | 15 | 20 |\n| 40-50 | 10 | 30 |\n| 50-60 | 8 | 38 |\n| 60-70 | 6 | 44 |\n| 70-80 | 2 | 46 |\n| **Total** | **N = 46** | |\n\n2.  **Find Quartiles:**\n    * **For $Q_1$**: Position is $(\\frac{46}{4})^{th} = 11.5^{th}$ item. $Q_1$ class is **30-40**.\n        $Q_1 = 30 + \\frac{11.5 - 5}{15} \\times 10 \\approx \\mathbf{34.33}$\n    * **For $Q_3$**: Position is $(\\frac{3 \\times 46}{4})^{th} = 34.5^{th}$ item. $Q_3$ class is **50-60**.\n        $Q_3 = 50 + \\frac{34.5 - 30}{8} \\times 10 = \\mathbf{55.625}$\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. Quartile Deviation (QD)** | $QD = \\frac{55.625 - 34.33}{2} \\approx \\mathbf{10.65}$ kg |\n| **2. Coefficient of QD** | $\\text{Coefficient} = \\frac{55.625 - 34.33}{55.625 + 34.33} \\approx \\mathbf{0.237}$ |\n\n***\n\n#### Question 12: Find the quartile deviation and its coefficient from the following data.\n\n| Class interval | Frequency |\n|:---:|:---:|\n| 20-30 | 8 |\n| 30-40 | 16 |\n| 40-50 | 4 |\n| 50-60 | 4 |\n| 60-70 | 3 |\n\n**Solution:**\n1.  **Cumulative Frequency Table:**\n\n| Class interval | Frequency (f) | c.f. |\n|:---:|:---:|:---:|\n| 20-30 | 8 | 8 |\n| 30-40 | 16 | 24 |\n| 40-50 | 4 | 28 |\n| 50-60 | 4 | 32 |\n| 60-70 | 3 | 35 |\n| **Total** | **N = 35** | |\n\n2.  **Find Quartiles:**\n    * **For $Q_1$**: Position is $(\\frac{35}{4})^{th} = 8.75^{th}$ item. $Q_1$ class is **30-40**.\n        $Q_1 = 30 + \\frac{8.75 - 8}{16} \\times 10 \\approx \\mathbf{30.47}$\n    * **For $Q_3$**: Position is $(\\frac{3 \\times 35}{4})^{th} = 26.25^{th}$ item. $Q_3$ class is **40-50**.\n        $Q_3 = 40 + \\frac{26.25 - 24}{4} \\times 10 = \\mathbf{45.625}$\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. Quartile Deviation (QD)** | $QD = \\frac{45.625 - 30.47}{2} \\approx \\mathbf{7.58}$ |\n| **2. Coefficient of QD** | $\\text{Coefficient} = \\frac{45.625 - 30.47}{45.625 + 30.47} \\approx \\mathbf{0.20}$ |\n\n***\n\n#### Question 13: Find the quartile deviation and its coefficient from the following data.\n\n| Size | Frequency |\n|:---:|:---:|\n| 4-8 | 6 |\n| 8-12 | 10 |\n| 12-16 | 18 |\n| 16-20 | 30 |\n| 20-24 | 15 |\n| 24-28 | 12 |\n| 28-32 | 10 |\n| 32-36 | 6 |\n| 36-40 | 5 |\n\n**Solution:**\n1.  **Find N and Quartile Positions:** $N = 112$.\n    * $Q_1$ position is $28^{th}$ item.\n    * $Q_3$ position is $84^{th}$ item.\n\n2.  **Find Quartiles from c.f. table:**\n    * **For $Q_1$**: c.f. is (6, 16, **34**,...). $Q_1$ class is **12-16**.\n        $Q_1 = 12 + \\frac{28 - 16}{18} \\times 4 \\approx \\mathbf{14.67}$\n    * **For $Q_3$**: c.f. is (..., 64, 79, **91**,...). $Q_3$ class is **24-28**.\n        $Q_3 = 24 + \\frac{84 - 79}{12} \\times 4 \\approx \\mathbf{25.67}$\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. Quartile Deviation (QD)** | $QD = \\frac{25.67 - 14.67}{2} = \\mathbf{5.5}$ |\n| **2. Coefficient of QD** | $\\text{Coefficient} = \\frac{25.67 - 14.67}{25.67 + 14.67} = \\frac{11}{40.34} \\approx \\mathbf{0.273}$ |\n\n***\n\n#### Question 14: Find the quartile deviation and its coefficient from the following data.\n\n| Height (inches) | No. of students |\n|:---:|:---:|\n| 60-62 | 4 |\n| 62-64 | 6 |\n| 64-66 | 8 |\n| 66-68 | 12 |\n| 68-70 | 7 |\n| 70-72 | 2 |\n\n**Solution:**\n1.  **Find N and Quartile Positions:** $N = 39$.\n    * $Q_1$ position is $9.75^{th}$ item.\n    * $Q_3$ position is $29.25^{th}$ item.\n\n2.  **Find Quartiles from c.f. table:**\n    * **For $Q_1$**: c.f. is (4, **10**,...). $Q_1$ class is **62-64**.\n        $Q_1 = 62 + \\frac{9.75 - 4}{6} \\times 2 \\approx \\mathbf{63.92}$\n    * **For $Q_3$**: c.f. is (..., 18, **30**,...). $Q_3$ class is **66-68**.\n        $Q_3 = 66 + \\frac{29.25 - 18}{12} \\times 2 \\approx \\mathbf{67.88}$\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. Quartile Deviation (QD)** | $QD = \\frac{67.88 - 63.92}{2} = \\mathbf{1.98}$ inches |\n| **2. Coefficient of QD** | $\\text{Coefficient} = \\frac{67.88 - 63.92}{67.88 + 63.92} = \\frac{3.96}{131.8} \\approx \\mathbf{0.03}$ |\n\n***\n\n#### Question 15: Find the quartile deviation and its coefficient from the following data.\n\n| Expenditure | No. of workers |\n|:---:|:---:|\n| 0-10 | 5 |\n| 10-20 | 15 |\n| 20-30 | 10 |\n| 30-40 | 8 |\n| 40-50 | 6 |\n\n**Solution:**\n1.  **Find N and Quartile Positions:** $N = 44$.\n    * $Q_1$ position is $11^{th}$ item.\n    * $Q_3$ position is $33^{rd}$ item.\n\n2.  **Find Quartiles from c.f. table:**\n    * **For $Q_1$**: c.f. is (5, **20**,...). $Q_1$ class is **10-20**.\n        $Q_1 = 10 + \\frac{11 - 5}{15} \\times 10 = \\mathbf{14}$\n    * **For $Q_3$**: c.f. is (..., 30, **38**,...). $Q_3$ class is **30-40**.\n        $Q_3 = 30 + \\frac{33 - 30}{8} \\times 10 = \\mathbf{33.75}$\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. Quartile Deviation (QD)** | $QD = \\frac{33.75 - 14}{2} = \\mathbf{9.875}$ |\n| **2. Coefficient of QD** | $\\text{Coefficient} = \\frac{33.75 - 14}{33.75 + 14} = \\frac{19.75}{47.75} \\approx \\mathbf{0.414}$ |\n\n***\n\n#### Question 16: Prepare a frequency distribution table (10-20, etc.) and find the QD and its coefficient.\n**Data:** 21, 45, 60, 57, 15, 41, 48, 50, 34, 29, 56, 40, 14, 62, 28, 70, 22, 30, 38, 74, 13, 47, 20, 53, 64, 34, 75, 66.\n\n**Solution:**\n1.  **Frequency Distribution Table:** ($N=28$)\n\n| Marks | Freq (f) | c.f. |\n|:---:|:---:|:---:|\n| 10-20 | 3 | 3 |\n| 20-30 | 5 | 8 |\n| 30-40 | 4 | 12 |\n| 40-50 | 5 | 17 |\n| 50-60 | 4 | 21 |\n| 60-70 | 4 | 25 |\n| 70-80 | 3 | 28 |\n\n2.  **Find Quartiles:**\n    * **For $Q_1$**: Position is $(\\frac{28}{4})^{th} = 7^{th}$ item. c.f. is (3, **8**,...). $Q_1$ class is **20-30**.\n        $Q_1 = 20 + \\frac{7 - 3}{5} \\times 10 = \\mathbf{28}$\n    * **For $Q_3$**: Position is $(\\frac{3 \\times 28}{4})^{th} = 21^{st}$ item. c.f. is exactly 21 at the end of the 50-60 class, so $Q_3$ is the upper boundary. $Q_3 = \\mathbf{60}$.\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. Quartile Deviation (QD)** | $QD = \\frac{60 - 28}{2} = \\mathbf{16}$ |\n| **2. Coefficient of QD** | $\\text{Coefficient} = \\frac{60 - 28}{60 + 28} = \\frac{32}{88} \\approx \\mathbf{0.364}$ |\n\n***\n\n#### Question 17: The following are the marks obtained by class 9 students. Prepare a frequency distribution table (10-20, etc.) and find the QD and its coefficient.\n**Data:** 22, 25, 46, 34, 57, 69, 44, 36, 12, 27, 50, 36, 35, 62, 46, 52, 54, 61, 66, 55, 29, 39, 40, 33, 14, 41, 25, 20, 16.\n\n**Solution:**\n1.  **Frequency Distribution Table:** ($N=29$)\n\n| Marks | Freq (f) | c.f. |\n|:---:|:---:|:---:|\n| 10-20 | 3 | 3 |\n| 20-30 | 6 | 9 |\n| 30-40 | 6 | 15 |\n| 40-50 | 5 | 20 |\n| 50-60 | 5 | 25 |\n| 60-70 | 4 | 29 |\n\n2.  **Find Quartiles:**\n    * **For $Q_1$**: Position is $(\\frac{29}{4})^{th} = 7.25^{th}$ item. c.f. is (3, **9**,...). $Q_1$ class is **20-30**.\n        $Q_1 = 20 + \\frac{7.25 - 3}{6} \\times 10 \\approx \\mathbf{27.08}$\n    * **For $Q_3$**: Position is $(\\frac{3 \\times 29}{4})^{th} = 21.75^{th}$ item. c.f. is (..., 20, **25**,...). $Q_3$ class is **50-60**.\n        $Q_3 = 50 + \\frac{21.75 - 20}{5} \\times 10 = \\mathbf{53.5}$\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. Quartile Deviation (QD)** | $QD = \\frac{53.5 - 27.08}{2} = \\mathbf{13.21}$ |\n| **2. Coefficient of QD** | $\\text{Coefficient} = \\frac{53.5 - 27.08}{53.5 + 27.08} = \\frac{26.42}{80.58} \\approx \\mathbf{0.328}$ |', 1, '2025-07-25 01:11:38', '2025-09-18 14:41:48');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(151, 14243, '\n# Grade 10 Compulsory Mathematics: Statistics Study Guide\n## Unit: Statistics | Chapter: Mean Deviation | Complete Solutions\n\n---\n\n## 🎯 Chapter Introduction: Understanding Mean Deviation\n\nThis guide focuses on **Mean Deviation**, another important tool for measuring the dispersion or spread of data. While quartile deviation only considers the middle 50% of data, mean deviation provides a more comprehensive picture by taking into account **every single data point**. It calculates the average distance of each data point from the center of the dataset (usually the mean or median). 📈\n\n### Key Learning Areas\n* **Defining Mean Deviation (MD):** Understanding what \"average deviation\" means.\n* **Coefficient of Mean Deviation:** Learning how to compare the variability of different datasets.\n* **Calculations from Mean:** Step-by-step methods to calculate MD using the arithmetic mean.\n* **Calculations from Median:** Step-by-step methods to calculate MD using the median.\n* **Choosing the Right Method:** Knowing when to use the mean versus the median.\n\n---\n\n## 🔑 Key Concepts & Definitions\n\nLet\'s break down the essential concepts before solving the problems.\n\n### 1. What is Mean Deviation?\n**Mean Deviation (MD)**, also called **Average Deviation**, is defined as the arithmetic average of the absolute differences (deviations) of each data point from a measure of central tendency. This central point can be the **mean**, **median**, or mode.\n\nIt essentially answers the question: \"On average, how far is each data point from the center?\" We use the **absolute value** (ignoring negative signs) because we only care about the distance, not the direction.\n\n### 2. Coefficient of Mean Deviation\nThe **Coefficient of Mean Deviation** is a **relative measure** of dispersion. Because MD is an absolute measure (expressed in the same units as the data, like \"marks\" or \"kg\"), its coefficient is necessary for comparing the variability of two different datasets. It is a pure number without units.\n\n* **Formula from Mean:**\n    $$\n    \\text{Coefficient of MD from Mean} = \\frac{\\text{MD from Mean}}{\\text{Mean}}\n    $$\n* **Formula from Median:**\n    $$\n    \\text{Coefficient of MD from Median} = \\frac{\\text{MD from Median}}{\\text{Median}}\n    $$\n\n### 3. Formulas & Calculation Methods\n\n#### A. Mean Deviation from Mean\n1.  **Calculate the Mean ($\\bar{X}$):** Find the average of the dataset ($\\bar{X} = \\frac{\\sum fx}{N}$).\n2.  **Find Deviations:** Subtract the mean from each data point or mid-value ($D = x - \\bar{X}$).\n3.  **Find Absolute Deviations:** Take the absolute value of each deviation ($|D|$).\n4.  **Calculate MD:** Find the average of these absolute deviations.\n    $$\n    \\text{MD from Mean} = \\frac{\\sum f|D|}{N} = \\frac{\\sum f|x - \\bar{X}|}{N}\n    $$\n\n#### B. Mean Deviation from Median\n1.  **Calculate the Median ($M_d$):** Find the middle value of the dataset.\n2.  **Find Deviations:** Subtract the median from each data point ($D = x - M_d$).\n3.  **Find Absolute Deviations:** Take the absolute value of each deviation ($|D|$).\n4.  **Calculate MD:** Find the average of these absolute deviations.\n    $$\n    \\text{MD from Median} = \\frac{\\sum f|D|}{N} = \\frac{\\sum f|x - M_d|}{N}\n    $$\n\n---\n\n## 📚 Exercise Solutions\n\n### Short Questions\n\n***\n\n#### Question 1: Define mean deviation.\n**Answer:**\nMean deviation is the average of the absolute values of the deviations of each item from a central point, which can be the mean, median, or mode.\n\n***\n\n#### Question 2: What is the coefficient of mean deviation?\n**Answer:**\nThe coefficient of mean deviation is a relative measure of dispersion used to compare the variability of two or more series. It is the ratio of the mean deviation to the average (mean or median) from which it was calculated.\n\n***\n\n#### Question 3: Write the formulas for mean deviation and its coefficient from the mean.\n**Answer:**\n* **MD from Mean:** $MD = \\frac{\\sum f|m - \\bar{X}|}{N}$\n* **Coefficient of MD from Mean:** $\\text{Coefficient} = \\frac{\\text{MD from Mean}}{\\text{Mean}}$\n\n***\n\n#### Question 4: Write the formulas for mean deviation and its coefficient from the median.\n**Answer:**\n* **MD from Median:** $MD = \\frac{\\sum f|m - M_d|}{N}$\n* **Coefficient of MD from Median:** $\\text{Coefficient} = \\frac{\\text{MD from Median}}{\\text{Median}}$\n\n***\n\n#### Question 5: Which method of calculating mean deviation is considered best and why?\n**Answer:**\nCalculating mean deviation from the **mean** is generally considered the best method. This is because the mean uses all the values in the dataset for its calculation, making it a more stable and representative measure of the center than the median, which is only a positional average.\n\n***\n\n#### Question 6: If $\\sum fm = 1000$, $N = 50$, and $\\sum f|m - \\bar{X}| = 308$, find the MD from mean and its coefficient.\n**Solution:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. Calculate Mean ($\\bar{X}$)** | $\\bar{X} = \\frac{\\sum fm}{N} = \\frac{1000}{50} = \\mathbf{20}$ |\n| **2. Calculate MD from Mean** | $MD = \\frac{\\sum f|m - \\bar{X}|}{N} = \\frac{308}{50} = \\mathbf{6.16}$ |\n| **3. Calculate Coefficient of MD** | $\\text{Coefficient} = \\frac{MD}{\\bar{X}} = \\frac{6.16}{20} = \\mathbf{0.308}$ |\n\n***\n\n#### Question 7: If $\\sum f|m - \\bar{X}| = 680$ and MD = 17, find $\\sum f$.\n**Solution:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **Find Total Frequency (N)** | $N = \\frac{\\sum f|m - \\bar{X}|}{MD} = \\frac{680}{17} = \\mathbf{40}$ |\n\n***\n\n#### Question 8: If Median ($M_d$) = 40, $\\sum f = 50$, and $\\sum f|m - M_d| = 530$, find MD from median and its coefficient.\n**Solution:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. Calculate MD from Median** | $MD = \\frac{\\sum f|m - M_d|}{N} = \\frac{530}{50} = \\mathbf{10.6}$ |\n| **2. Calculate Coefficient of MD** | $\\text{Coefficient} = \\frac{MD}{M_d} = \\frac{10.6}{40} = \\mathbf{0.265}$ |\n\n***\n\n#### Question 9: If the coefficient of MD from median is 0.5 and the median is 40, find the MD.\n**Solution:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **Find Mean Deviation (MD)** | $MD = \\text{Coefficient} \\times M_d = 0.5 \\times 40 = \\mathbf{20}$ |\n\n---\n\n### Long Questions (Worked Examples)\n\n***\n\n#### Question 10: Find the MD and its coefficient from the median for the data: 10, 50, 60, 40, 30, 20.\n**Solution:**\n1.  **Arrange data and find Median:**\n    * Data: 10, 20, 30, 40, 50, 60.\n    * $M_d = \\frac{30 + 40}{2} = 35$.\n\n2.  **Data Table:**\n\n| X | D = &#124;X - 35&#124; |\n|:---:|:---:|\n| 10 | 25 |\n| 20 | 15 |\n| 30 | 5 |\n| 40 | 5 |\n| 50 | 15 |\n| 60 | 25 |\n| **Total** | **Σ&#124;D&#124; = 90** |\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. MD from Median** | $MD = \\frac{\\sum|D|}{N} = \\frac{90}{6} = \\mathbf{15}$ |\n| **2. Coefficient of MD** | $\\text{Coefficient} = \\frac{MD}{M_d} = \\frac{15}{35} \\approx \\mathbf{0.429}$ |\n\n***\n\n#### Question 11: Find the MD and its coefficient from the median.\n\n| Marks (X) | Freq (f) |\n|:---:|:---:|\n| 20 | 2 |\n| 30 | 4 |\n| 40 | 5 |\n| 50 | 7 |\n| 60 | 2 |\n| 70 | 1 |\n\n**Solution:**\n1.  **Find Median:** $N=21$. The median position is the $11^{th}$ item. From c.f. (2, 6, **11**,...), the $M_d = \\mathbf{40}$.\n\n2.  **Data Table:**\n\n| X | f | D = &#124;X - 40&#124; | f&#124;D&#124; |\n|:---:|:---:|:---:|:---:|\n| 20 | 2 | 20 | 40 |\n| 30 | 4 | 10 | 40 |\n| 40 | 5 | 0 | 0 |\n| 50 | 7 | 10 | 70 |\n| 60 | 2 | 20 | 40 |\n| 70 | 1 | 30 | 30 |\n| **Total** | **N=21** | | **Σf&#124;D&#124;=220** |\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. MD from Median** | $MD = \\frac{\\sum f|D|}{N} = \\frac{220}{21} \\approx \\mathbf{10.48}$ |\n| **2. Coefficient of MD** | $\\text{Coefficient} = \\frac{MD}{M_d} = \\frac{10.48}{40} \\approx \\mathbf{0.262}$ |\n\n***\n\n#### Question 12: Find the MD and its coefficient from the mean.\n\n| Marks (X) | Freq (f) |\n|:---:|:---:|\n| 20 | 2 |\n| 25 | 5 |\n| 30 | 6 |\n| 35 | 5 |\n| 40 | 2 |\n\n**Solution:**\n1.  **Find Mean:** $N=20$. $\\sum fX = 600$. $\\bar{X} = \\frac{600}{20} = \\mathbf{30}$.\n\n2.  **Data Table:**\n\n| X | f | D = &#124;X - 30&#124; | f&#124;D&#124; |\n|:---:|:---:|:---:|:---:|\n| 20 | 2 | 10 | 20 |\n| 25 | 5 | 5 | 25 |\n| 30 | 6 | 0 | 0 |\n| 35 | 5 | 5 | 25 |\n| 40 | 2 | 10 | 20 |\n| **Total** | **N=20** | | **Σf&#124;D&#124;=90** |\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. MD from Mean** | $MD = \\frac{\\sum f|D|}{N} = \\frac{90}{20} = \\mathbf{4.5}$ |\n| **2. Coefficient of MD** | $\\text{Coefficient} = \\frac{MD}{\\bar{X}} = \\frac{4.5}{30} = \\mathbf{0.15}$ |\n\n***\n\n#### Question 13: Calculate the MD from the median for the following data.\n\n| Marks | Freq (f) |\n|:---:|:---:|\n| 10-20 | 5 |\n| 20-30 | 4 |\n| 30-40 | 5 |\n| 40-50 | 4 |\n| 50-60 | 2 |\n\n**Solution:**\n1.  **Find Median:** $N=20$. Median class is **30-40**.\n    $M_d = L + \\frac{\\frac{N}{2} - c.f.}{f} \\times i = 30 + \\frac{10 - 9}{5} \\times 10 = \\mathbf{32}$.\n\n2.  **Data Table:**\n\n| Marks | m | f | D = &#124;m - 32&#124; | f&#124;D&#124; |\n|:---:|:---:|:---:|:---:|:---:|\n| 10-20 | 15 | 5 | 17 | 85 |\n| 20-30 | 25 | 4 | 7 | 28 |\n| 30-40 | 35 | 5 | 3 | 15 |\n| 40-50 | 45 | 4 | 13 | 52 |\n| 50-60 | 55 | 2 | 23 | 46 |\n| **Total** | | **N=20** | | **Σf&#124;D&#124;=226** |\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **MD from Median** | $MD = \\frac{\\sum f|D|}{N} = \\frac{226}{20} = \\mathbf{11.3}$ |\n\n***\n\n#### Question 14: Compute the mean deviation and its coefficient from the median.\n\n| Marks (X) | Freq (f) |\n|:---:|:---:|\n| 55 | 4 |\n| 65 | 5 |\n| 75 | 6 |\n| 85 | 3 |\n| 95 | 2 |\n\n**Solution:**\n1.  **Find Median:** $N=20$. The median is the $10.5^{th}$ item. From c.f. (4, 9, **15**,...), the $M_d = \\mathbf{75}$.\n\n2.  **Data Table:**\n\n| X | f | D = &#124;X - 75&#124; | f&#124;D&#124; |\n|:---:|:---:|:---:|:---:|\n| 55 | 4 | 20 | 80 |\n| 65 | 5 | 10 | 50 |\n| 75 | 6 | 0 | 0 |\n| 85 | 3 | 10 | 30 |\n| 95 | 2 | 20 | 40 |\n| **Total** | **N=20** | | **Σf&#124;D&#124;=200** |\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. MD from Median** | $MD = \\frac{\\sum f|D|}{N} = \\frac{200}{20} = \\mathbf{10}$ |\n| **2. Coefficient of MD** | $\\text{Coefficient} = \\frac{MD}{M_d} = \\frac{10}{75} \\approx \\mathbf{0.133}$ |\n\n***\n\n#### Question 15: Find the MD and its coefficient from the median.\n\n| Marks (X) | Freq (f) |\n|:---:|:---:|\n| 10 | 2 |\n| 20 | 3 |\n| 30 | 9 |\n| 40 | 21 |\n| 50 | 11 |\n| 60 | 5 |\n\n**Solution:**\n1.  **Find Median:** $N=51$. The median is the $26^{th}$ item. From c.f. (2, 5, 14, **35**,...), the $M_d = \\mathbf{40}$.\n\n2.  **Data Table:**\n\n| X | f | D = &#124;X - 40&#124; | f&#124;D&#124; |\n|:---:|:---:|:---:|:---:|\n| 10 | 2 | 30 | 60 |\n| 20 | 3 | 20 | 60 |\n| 30 | 9 | 10 | 90 |\n| 40 | 21 | 0 | 0 |\n| 50 | 11 | 10 | 110 |\n| 60 | 5 | 20 | 100 |\n| **Total** | **N=51** | | **Σf&#124;D&#124;=420** |\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **1. MD from Median** | $MD = \\frac{\\sum f|D|}{N} = \\frac{420}{51} \\approx \\mathbf{8.24}$ |\n| **2. Coefficient of MD** | $\\text{Coefficient} = \\frac{MD}{M_d} = \\frac{8.24}{40} \\approx \\mathbf{0.206}$ |\n\n***\n\n#### Question 16: Find the mean deviation from the median for the unsorted data below.\n\n| Marks (X) | Freq (f) |\n|:---:|:---:|\n| 20 | 3 |\n| 10 | 4 |\n| 30 | 5 |\n| 50 | 2 |\n| 40 | 1 |\n\n**Solution:**\n1.  **Sort Data and Find Median:** $N=15$. The median is the $8^{th}$ item.\n\n| X (sorted) | f | c.f. |\n|:---:|:---:|:---:|\n| 10 | 4 | 4 |\n| 20 | 3 | 7 |\n| 30 | 5 | **12** |\n| 40 | 1 | 13 |\n| 50 | 2 | 15 |\n\nThe 8th item falls in the class with mark **30**. So, $M_d = \\mathbf{30}$.\n\n2.  **Data Table:**\n\n| X | f | D = &#124;X - 30&#124; | f&#124;D&#124; |\n|:---:|:---:|:---:|:---:|\n| 10 | 4 | 20 | 80 |\n| 20 | 3 | 10 | 30 |\n| 30 | 5 | 0 | 0 |\n| 40 | 1 | 10 | 10 |\n| 50 | 2 | 20 | 40 |\n| **Total** | **N=15** | | **Σf&#124;D&#124;=160** |\n\n3.  **Final Calculation:**\n\n| Calculation Step | Formula & Result |\n|:---|:---|\n| **MD from Median** | $MD = \\frac{\\sum f|D|}{N} = \\frac{160}{15} \\approx \\mathbf{10.67}$ |', 1, '2025-07-25 01:11:38', '2025-09-18 14:38:40'),
(152, 14244, '\n# Grade 10 Compulsory Mathematics: Statistics Study Guide\n## Unit: Statistics | Chapter: Standard Deviation & Coefficient of Variation | Complete Solutions\n\n---\n\n## 🎯 Chapter Introduction: Standard Deviation\n\nWelcome to the complete guide on **Standard Deviation**, the most widely used and robust measure of dispersion in statistics. While other measures like Quartile and Mean Deviation provide valuable insights, Standard Deviation is considered the most reliable because it gives more weight to the data points that are farther from the mean. It\'s the cornerstone for many advanced statistical analyses. 🔬\n\n### Key Learning Areas\n* **Defining Standard Deviation ($\\sigma$):** Understanding the \"root mean square deviation.\"\n* **Coefficient of Standard Deviation:** A relative measure for comparing spread.\n* **Coefficient of Variation (C.V.):** A standardized measure of dispersion, expressed as a percentage, used to compare consistency.\n* **Calculation Methods:** Step-by-step processes for individual, discrete, and continuous series.\n* **Consistency and Stability:** Using C.V. to determine which dataset is more consistent.\n\n---\n\n## 🔑 Key Concepts & Definitions\n\nLet\'s dive into the foundational ideas.\n\n### 1. What is Standard Deviation?\n**Standard Deviation ($\\sigma$)** is the **positive square root of the mean of the squared deviations from the arithmetic mean**. This might sound complex, but it\'s a very logical process:\n\n1.  **Find the Mean ($\\bar{X}$):** Calculate the average of the data.\n2.  **Find Deviations:** Find the difference between each data point and the mean.\n3.  **Square the Deviations:** Square each difference. This makes all values positive and gives more weight to larger deviations.\n4.  **Find the Average of Squares:** Calculate the mean of these squared deviations. This value is called the **Variance ($\\sigma^2$)**.\n5.  **Take the Square Root:** The square root of the variance is the **Standard Deviation**.\n\nIt is also known as the **root mean square deviation** and is denoted by the Greek letter **$\\sigma$** (sigma).\n\n### 2. Coefficient of Standard Deviation\nThis is a **relative measure** of dispersion. It\'s the ratio of the Standard Deviation to the arithmetic mean, which allows for a standardized comparison of variability between different datasets.\n\n* **Formula:**\n    $$\n    \\text{Coefficient of SD} = \\frac{\\sigma}{\\bar{X}}\n    $$\n\n### 3. Coefficient of Variation (C.V.)\nThe **Coefficient of Variation (C.V.)** is the most common relative measure of dispersion. It\'s simply the Coefficient of Standard Deviation expressed as a **percentage**.\n\n* **Formula:**\n    $$\n    C.V. = \\frac{\\sigma}{\\bar{X}} \\times 100\\%\n    $$\n\n**Why is C.V. important?** It\'s used to compare the consistency or stability of different datasets.\n* A **low C.V.** means the data points are close to the mean, indicating **high consistency** or stability.\n* A **high C.V.** means the data points are spread out, indicating **low consistency**.\n\n---\n\n## 📚 Exercise Solutions\n\n### Short Questions\n\n***\n\n#### Question 1: Define standard deviation.\n**Answer:**\nStandard deviation is the positive square root of the mean of the squared deviations of data points from their arithmetic mean. It\'s often called the root mean square deviation and is denoted by $\\sigma$.\n\n***\n\n#### Question 2: What is the coefficient of standard deviation and its formula?\n**Answer:**\nThe coefficient of standard deviation is a relative measure of dispersion, calculated as the ratio of the standard deviation ($\\sigma$) to the mean ($\\bar{X}$). Its formula is:\n$$\n\\text{Coefficient of SD} = \\frac{\\sigma}{\\bar{X}}\n$$\n\n***\n\n#### Question 3: Define the coefficient of variation and write its formula.\n**Answer:**\nThe coefficient of variation (C.V.) is the coefficient of standard deviation expressed as a percentage. It is a unit-less measure used for comparing the consistency of data.\n$$\nC.V. = \\frac{\\sigma}{\\bar{X}} \\times 100\\%\n$$\n\n***\n\n#### Question 4: What does a low coefficient of variation mean?\n**Answer:**\nA low coefficient of variation indicates that the data points are clustered closely around the mean. This signifies that the data is more **consistent**, uniform, stable, or homogenous.\n\n***\n\n#### Question 5: What is the difference between standard deviation and mean deviation?\n**Answer:**\n**Standard deviation** is the square root of the average of *squared* deviations from the mean. **Mean deviation** is the average of the *absolute* deviations from the mean or median. Standard deviation gives greater weight to extreme values because it squares the deviations.\n\n***\n\n#### Question 6: What is the difference between the coefficient of standard deviation and the coefficient of variation?\n**Answer:**\nThe coefficient of standard deviation is the ratio $\\frac{\\sigma}{\\bar{X}}$. The coefficient of variation is simply this same ratio multiplied by 100 to express it as a percentage ($C.V. = \\frac{\\sigma}{\\bar{X}} \\times 100\\%$).\n\n***\n\n#### Question 7: If $\\sum f(m - \\bar{X})^2 = 484$, $N = 24$, and $\\bar{X} = 25$, find the standard deviation and its coefficient.\n**Solution:**\n1.  **Standard Deviation ($\\sigma$):**\n    $$\n    \\sigma = \\sqrt{\\frac{\\sum f(m - \\bar{X})^2}{N}} = \\sqrt{\\frac{484}{24}} \\approx \\sqrt{20.167} \\approx 4.49\n    $$\n2.  **Coefficient of SD:**\n    $$\n    \\text{Coefficient} = \\frac{\\sigma}{\\bar{X}} = \\frac{4.49}{25} \\approx 0.18\n    $$\n\n***\n\n#### Question 8: If $\\sum fd^2 = 848$, $\\sum fd = 0$, $N=100$, and Assumed Mean A=12, find the standard deviation.\n**Solution:**\nSince $\\sum fd = 0$, the assumed mean (A) is equal to the actual mean ($\\bar{X}$).\n$$\n\\sigma = \\sqrt{\\frac{\\sum fd^2}{N} - \\left(\\frac{\\sum fd}{N}\\right)^2} = \\sqrt{\\frac{848}{100} - \\left(\\frac{0}{100}\\right)^2} = \\sqrt{8.48} \\approx 2.91\n$$\n\n***\n\n#### Question 9: If $\\sum fd\'^2 = 125$, $\\sum fd\' = -4$, $N=20$, and class size $C=10$, find the standard deviation.\n**Solution:**\nUsing the step-deviation formula:\n$$\n\\sigma = \\sqrt{\\frac{\\sum fd\'^2}{N} - \\left(\\frac{\\sum fd\'}{N}\\right)^2} \\times C = \\sqrt{\\frac{125}{20} - \\left(\\frac{-4}{20}\\right)^2} \\times 10\n$$\n$$\n\\sigma = \\sqrt{6.25 - (-0.2)^2} \\times 10 = \\sqrt{6.25 - 0.04} \\times 10 = \\sqrt{6.21} \\times 10 \\approx 2.492 \\times 10 \\approx 24.92\n$$\n\n---\n\n### Long Questions (Worked Examples)\n\n***\n\n#### Question 10: Find the standard deviation and its coefficient for the data: 5, 10, 15, 20, 25.\n**Solution:**\n1.  **Find Mean:** $N=5$. $\\sum X = 5+10+15+20+25 = 75$.\n    $\\bar{X} = \\frac{75}{5} = 15$.\n2.  **Calculations:**\n| X | $D = X - \\bar{X}$ | $D^2$ |\n|:---:|:---:|:---:|\n| 5 | -10 | 100 |\n| 10 | -5 | 25 |\n| 15 | 0 | 0 |\n| 20 | 5 | 25 |\n| 25 | 10 | 100 |\n| **Total** | | **$\\sum D^2 = 250$** |\n\n* **Standard Deviation ($\\sigma$)** = $\\sqrt{\\frac{\\sum D^2}{N}} = \\sqrt{\\frac{250}{5}} = \\sqrt{50} \\approx \\mathbf{7.07}$\n* **Coefficient of SD** = $\\frac{\\sigma}{\\bar{X}} = \\frac{7.07}{15} \\approx \\mathbf{0.47}$\n\n***\n\n#### Question 11: Find the standard deviation and C.V. for the data: 30, 33, 37, 40, 44, 50.\n**Solution:**\n1.  **Find Mean:** $N=6$. $\\sum X = 30+33+37+40+44+50 = 234$.\n    $\\bar{X} = \\frac{234}{6} = 39$.\n2.  **Calculations:**\n\n| X | $D = X - 39$ | $D^2$ |\n|:---:|:---:|:---:|\n| 30 | -9 | 81 |\n| 33 | -6 | 36 |\n| 37 | -2 | 4 |\n| 40 | 1 | 1 |\n| 44 | 5 | 25 |\n| 50 | 11 | 121 |\n| **Total**| | **$\\sum D^2 = 268$** |\n\n* **Standard Deviation ($\\sigma$)** = $\\sqrt{\\frac{268}{6}} = \\sqrt{44.67} \\approx \\mathbf{6.68}$\n* **Coefficient of Variation (C.V.)** = $\\frac{6.68}{39} \\times 100\\% \\approx \\mathbf{17.13\\%}$\n\n***\n\n#### Question 12: Find the standard deviation from the given data.\n\n| x | f |\n|:---:|:---:|\n| 6 | 1 |\n| 8 | 14 |\n| 10 | 25 |\n| 12 | 27 |\n| 14 | 18 |\n| 16 | 9 |\n| 18 | 4 |\n| 20 | 2 |\n\n**Solution:**\n1.  **Find Mean ($\\bar{X}$):** N=100. $\\sum fx = (6 \\times 1) + (8 \\times 14) + ... + (20 \\times 2) = 1200$.\n    $\\bar{X} = \\frac{1200}{100} = 12$.\n2.  **Calculations for SD:**\n\n| x | f | $D = x-12$ | $D^2$ | $fD^2$ |\n|:---:|:---:|:---:|:---:|:---:|\n| 6 | 1 | -6 | 36 | 36 |\n| 8 | 14 | -4 | 16 | 224 |\n| 10 | 25 | -2 | 4 | 100 |\n| 12 | 27 | 0 | 0 | 0 |\n| 14 | 18 | 2 | 4 | 72 |\n| 16 | 9 | 4 | 16 | 144 |\n| 18 | 4 | 6 | 36 | 144 |\n| 20 | 2 | 8 | 64 | 128 |\n| **Total** | **N=100** | | | **$\\sum fD^2=848$** |\n\n* **Standard Deviation ($\\sigma$)** = $\\sqrt{\\frac{\\sum fD^2}{N}} = \\sqrt{\\frac{848}{100}} = \\sqrt{8.48} \\approx \\mathbf{2.91}$\n\n***\n\n#### Question 13: Calculate the coefficient of variation from the data given below.\n\n| Class Interval | Frequency |\n|:---:|:---:|\n| 0-10 | 5 |\n| 10-20 | 8 |\n| 20-30 | 15 |\n| 30-40 | 16 |\n| 40-50 | 6 |\n\n**Solution:**\n1.  **Find Mean ($\\bar{X}$):** N=50. $\\sum fm = 1350$.\n    $\\bar{X} = \\frac{1350}{50} = 27$.\n2.  **Find Standard Deviation ($\\sigma$):**\n\n| m | f | $D = m - 27$ | $D^2$ | $fD^2$ |\n|:---:|:---:|:---:|:---:|:---:|\n| 5 | 5 | -22 | 484 | 2420 |\n| 15 | 8 | -12 | 144 | 1152 |\n| 25 | 15 | -2 | 4 | 60 |\n| 35 | 16 | 8 | 64 | 1024 |\n| 45 | 6 | 18 | 324 | 1944 |\n| **Total** | **N=50** | | | **$\\sum fD^2=6600$** |\n    $\\sigma = \\sqrt{\\frac{\\sum fD^2}{N}} = \\sqrt{\\frac{6600}{50}} = \\sqrt{132} \\approx 11.49$.\n4.  **Calculate C.V.:**\n    **C.V.** = $\\frac{\\sigma}{\\bar{X}} \\times 100\\% = \\frac{11.49}{27} \\times 100\\% \\approx \\mathbf{42.56\\%}$\n\n***\n\n#### Question 14: Calculate the coefficient of variation from the data given below.\n\n| Class Interval | Frequency |\n|:---:|:---:|\n| 0-20 | 2 |\n| 20-40 | 3 |\n| 40-60 | 4 |\n| 60-80 | 5 |\n| 80-100 | 6 |\n\n**Solution:**\n1.  **Find Mean ($\\bar{X}$):** N=20.\n2. \n| C.I. | m | f | fm |\n|:---:|:---:|:---:|:---:|\n| 0-20 | 10 | 2 | 20 |\n| 20-40 | 30 | 3 | 90 |\n| 40-60 | 50 | 4 | 200 |\n| 60-80 | 70 | 5 | 350 |\n| 80-100 | 90 | 6 | 540 |\n| **Total** | | **N=20** | **$\\sum fm = 1200$** |\n\n    $\\bar{X} = \\frac{1200}{20} = 60$.\n    \n3.  **Find Standard Deviation ($\\sigma$):**\n| m | f | $D=m-60$ | $D^2$ | $fD^2$ |\n|:---:|:---:|:---:|:---:|:---:|\n| 10 | 2 | -50 | 2500 | 5000 |\n| 30 | 3 | -30 | 900 | 2700 |\n| 50 | 4 | -10 | 100 | 400 |\n| 70 | 5 | 10 | 100 | 500 |\n| 90 | 6 | 30 | 900 | 5400 |\n| **Total** | **N=20** | | | **$\\sum fD^2 = 14000$** |\n    $\\sigma = \\sqrt{\\frac{14000}{20}} = \\sqrt{700} \\approx 26.46$.\n4.  **Calculate C.V.:**\n    **C.V.** = $\\frac{26.46}{60} \\times 100\\% \\approx \\mathbf{44.1\\%}$\n\n***\n\n#### Question 15: Find the standard deviation of the data given below.\n\n| Class Interval | Frequency |\n|:---:|:---:|\n| 5-15 | 7 |\n| 15-25 | 3 |\n| 25-35 | 6 |\n| 35-45 | 4 |\n| 45-55 | 5 |\n\n**Solution:**\n\n1.  **Find Mean ($\\bar{X}$):** N=25. $\\sum fm = (10\\times7) + (20\\times3) + (30\\times6) + (40\\times4) + (50\\times5) = 720$.\n    $\\bar{X} = \\frac{720}{25} = 28.8$.\n2.  **Calculations for SD:**\n\n| m | f | $D = m-28.8$ | $D^2$ | $fD^2$ |\n|:---:|:---:|:---:|:---:|:---:|\n| 10 | 7 | -18.8 | 353.44 | 2474.08 |\n| 20 | 3 | -8.8 | 77.44 | 232.32 |\n| 30 | 6 | 1.2 | 1.44 | 8.64 |\n| 40 | 4 | 11.2 | 125.44 | 501.76 |\n| 50 | 5 | 21.2 | 449.44 | 2247.2 |\n| **Total** | **N=25** | | | **$\\sum fD^2 = 5464$** |\n* **Standard Deviation ($\\sigma$)** = $\\sqrt{\\frac{5464}{25}} = \\sqrt{218.56} \\approx \\mathbf{14.78}$\n\n***\nThis guide now contains a comprehensive set of examples. Practice these thoroughly to master the concepts of Standard Deviation and Coefficient of Variation. Good luck! 🌟\n', 1, '2025-07-25 01:11:38', '2025-09-18 14:27:45'),
(153, 14246, '# Conic Sections: A Guide to Slicing a Cone\r\n## An Overview of Circles, Ellipses, Parabolas, and Hyperbolas\r\n\r\n---\r\n\r\n## 🎯 Introduction: The Carrot and the Knife\r\n\r\nHave you ever wondered about the different shapes you can get by slicing a carrot (which is shaped like a cone) with a knife? Depending on the angle of your cut, you can create a circle, an oval (ellipse), and other interesting curves. These shapes are known as **conic sections**. They are fundamental curves in geometry and appear everywhere in science and nature. 🥕\r\n\r\nThis guide explains how these four key shapes are formed by slicing a cone.\r\n\r\n---\r\n\r\n## 🔑 What is a Cone?\r\n\r\nIn geometry, when we talk about a cone for conic sections, we\'re usually referring to a **double-napped right circular cone**.\r\n\r\nImagine two fixed lines, AB and CD, that intersect at a point **O**. If you rotate the line AB around the fixed line CD, the surface you create is this double cone.\r\n\r\n![Diagram of a double-napped right circular cone](https://seekoguide.com/images/conic/1.svg)\r\n\r\nHere are the key parts:\r\n* **Vertex (O):** The point where the two cones meet.\r\n* **Axis (CD):** The fixed line that the other line rotates around.\r\n* **Generator (AB):** The moving line that traces the surface of the cone.\r\n* **Nappes:** The two separate cones are called nappes (an upper nappe and a lower nappe).\r\n\r\n---\r\n\r\n## 🔪 Creating Shapes by Slicing the Cone\r\n\r\nThe four conic sections are created by slicing this double cone with a flat plane at different angles.\r\n\r\n### 1. The Circle\r\nWhen you slice the cone with a plane that is **perfectly horizontal** (perpendicular to the axis) and parallel to the circular base, the resulting shape is a **circle**.\r\n\r\n![Formation of a circle by slicing a cone](https://seekoguide.com/images/conic/circle.svg)\r\n\r\n---\r\n\r\n### 2. The Ellipse\r\nIf you tilt the plane slightly, so the slice is **not parallel to the base** but still cuts completely through only one nappe, the resulting shape is an **ellipse**. An ellipse is essentially a stretched or squashed circle.\r\n\r\n![Formation of an ellipse by slicing a cone](https://seekoguide.com/images/conic/ellipse.svg)\r\n\r\n---\r\n\r\n### 3. The Parabola\r\nWhen you slice the cone with a plane that is **parallel to one of the generator lines** (the slanted edge of the cone), the curve you get is a **parabola**. This curve does not close back on itself.\r\n\r\n![Formation of a parabola by slicing a cone](https://seekoguide.com/images/conic/parabola.svg)\r\n\r\n---\r\n\r\n### 4. The Hyperbola\r\nIf you slice the cone with a plane that is steep enough (parallel to the axis or at an even steeper angle) to cut through **both the upper and lower nappes**, you get a **hyperbola**. A hyperbola consists of two separate, mirror-image curves.\r\n\r\n![Formation of a hyperbola by slicing a cone](https://seekoguide.com/images/conic/hyperbola.svg)\r\n\r\n---\r\n\r\n## ✍️ Exercises\r\n\r\n### Question 1: Parts of a Cone\r\nName the different parts of the cone indicated by the letters a, b, c, d, and e in the figure below.\r\n\r\n![Cone with labeled parts for exercise 1](https://seekoguide.com/images/conic/one.svg)\r\n\r\n**Answer:**\r\n* **a:** Axis\r\n* **b:** Upper Nappe\r\n* **c:** Lower Nappe\r\n* **d:** Generator\r\n* **e:** Vertex\r\n\r\n---\r\n\r\n### Question 2: Identify the Conic Section\r\nName the conic section shown in each of the following figures.\r\n\r\n1.  ![Diagram for exercise 2a](https://seekoguide.com/images/conic/a.svg)\r\n    **Answer:** Ellipse\r\n\r\n2.  ![Diagram for exercise 2b](https://seekoguide.com/images/conic/b.svg)\r\n    **Answer:** Parabola\r\n\r\n3.  ![Diagram for exercise 2c](https://seekoguide.com/images/conic/c.svg)\r\n    **Answer:** Circle\r\n\r\n4.  ![Diagram for exercise 2d](https://seekoguide.com/images/conic/d.svg)\r\n    **Answer:** Hyperbola', 1, '2025-07-25 01:11:38', '2025-09-18 14:27:45');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(154, 13120, '# Transformations: Complete Study Guide for Combined Transformations\r\n## A Comprehensive Guide to Reflection, Rotation, Translation, and Enlargement\r\n\r\n---\r\n\r\n## 🎯 Introduction: Master Combined Transformations\r\n\r\nWelcome to the ultimate guide on **Combined Transformations in Coordinate Geometry**. This comprehensive study guide covers everything you need to know about performing multiple transformations in sequence. When we combine transformations, we perform **two or more transformations one after another** on geometric shapes or points.\r\n\r\nUnderstanding combined transformations is essential for:\r\n- Solving complex coordinate geometry problems\r\n- Understanding geometric patterns and symmetries\r\n- Preparing for advanced mathematics examinations\r\n- Developing spatial reasoning skills\r\n\r\n**Key Learning Objective**: Master the art of applying multiple transformations systematically and understand how the order of operations affects the final result. 📐\r\n\r\n---\r\n\r\n## 🔑 Essential Transformation Rules\r\n\r\n### 1. Reflection Transformations\r\n**Basic Reflection Rules:**\r\n\r\n* **Reflection in the x-axis (y = 0):**\r\n  - Rule: $P(x, y) \\rightarrow P\'(x, -y)$\r\n  - Effect: Point flips across horizontal axis\r\n\r\n* **Reflection in the y-axis (x = 0):**\r\n  - Rule: $P(x, y) \\rightarrow P\'(-x, y)$\r\n  - Effect: Point flips across vertical axis\r\n\r\n* **Reflection in the line y = x:**\r\n  - Rule: $P(x, y) \\rightarrow P\'(y, x)$\r\n  - Effect: Coordinates interchange\r\n\r\n* **Reflection in the line y = -x (or x + y = 0):**\r\n  - Rule: $P(x, y) \\rightarrow P\'(-y, -x)$\r\n  - Effect: Coordinates interchange and both change sign\r\n\r\n* **Reflection in vertical line x = k:**\r\n  - Rule: $P(x, y) \\rightarrow P\'(2k - x, y)$\r\n  - Effect: Point reflects across vertical line\r\n\r\n* **Reflection in horizontal line y = k:**\r\n  - Rule: $P(x, y) \\rightarrow P\'(x, 2k - y)$\r\n  - Effect: Point reflects across horizontal line\r\n\r\n### 2. Rotation Transformations (about origin)\r\n**Standard Rotation Rules:**\r\n\r\n* **Rotation by +90° (Quarter-turn anticlockwise):**\r\n  - Rule: $P(x, y) \\rightarrow P\'(-y, x)$\r\n  - Memory tip: \"Negative y becomes new x, old x becomes new y\"\r\n\r\n* **Rotation by -90° (Quarter-turn clockwise):**\r\n  - Rule: $P(x, y) \\rightarrow P\'(y, -x)$\r\n  - Memory tip: \"y becomes new x, negative old x becomes new y\"\r\n\r\n* **Rotation by 180° (Half-turn):**\r\n  - Rule: $P(x, y) \\rightarrow P\'(-x, -y)$\r\n  - Memory tip: \"Both coordinates change sign\"\r\n\r\n### 3. Translation Transformations\r\n**Vector Translation Rule:**\r\n* **Translation by vector $\\begin{pmatrix} a \\\\ b \\end{pmatrix}$:**\r\n  - Rule: $P(x, y) \\rightarrow P\'(x+a, y+b)$\r\n  - Effect: Point moves a units horizontally, b units vertically\r\n\r\n### 4. Enlargement Transformations\r\n**Scaling Rules:**\r\n* **Enlargement from origin with scale factor k:**\r\n  - Rule: $P(x, y) \\rightarrow P\'(kx, ky)$\r\n  - Effect: Distance from origin multiplied by k\r\n\r\n* **Enlargement from center (a,b) with scale factor k:**\r\n  - Rule: $P(x, y) \\rightarrow P\'(k(x-a)+a, k(y-b)+b)$\r\n  - Effect: Point moves relative to center, scaled by k\r\n\r\n---\r\n\r\n## 📚 Worked Examples and Solutions\r\n\r\n### Problem 1: Triangle Translation and Enlargement\r\n**Question:** A triangle ABC with vertices A(2, 3), B(2, 6) and C(3, 4) is translated by T = $\\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}$ and the image so obtained is enlarged by E[(0, 0), 2]. Write the co-ordinates of the vertices of the images so formed and represent △ABC and its both images in the same graph.\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply translation**\r\nTranslation rule: $P(x, y) \\rightarrow P\'(x-2, y+3)$\r\n- A(2, 3) → A\'(0, 6)\r\n- B(2, 6) → B\'(0, 9) \r\n- C(3, 4) → C\'(1, 7)\r\n\r\n**Step 2: Apply enlargement**\r\nEnlargement rule: $P(x, y) \\rightarrow P\'\'(2x, 2y)$\r\n- A\'(0, 6) → A\'\'(0, 12)\r\n- B\'(0, 9) → B\'\'(0, 18)\r\n- C\'(1, 7) → C\'\'(2, 14)\r\n\r\n**Final Answer:**\r\n- After translation: **A\'(0, 6), B\'(0, 9), C\'(1, 7)**\r\n- After enlargement: **A\'\'(0, 12), B\'\'(0, 18), C\'\'(2, 14)**\r\n\r\n![Graph of Triangle ABC and its transformations](https://seekoguide.com/images/comb-of-transformation/transformation.svg)\r\n\r\n### Problem 2: Combined Transformation with Composition Notation\r\n**Question:** E denotes the enlargement about the centre (3, 1) with a scale factor of 2 and R denotes a reflection on the line y = x. Find the image of △ABC having the vertices A(2, 3), B(4, 5) and C(1, -2) under the combined transformation E∘R. Draw both figure on the same graph paper.\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply R (Reflection on y = x)**\r\nReflection rule: $P(x, y) \\rightarrow P\'(y, x)$\r\n- A(2, 3) → A\'(3, 2)\r\n- B(4, 5) → B\'(5, 4) \r\n- C(1, -2) → C\'(-2, 1)\r\n\r\n**Step 2: Apply E (Enlargement from center (3, 1) with scale factor 2)**\r\nEnlargement rule: $P(x, y) \\rightarrow P\'\'(2(x-3)+3, 2(y-1)+1)$\r\n- A\'(3, 2) → A\'\'(2(3-3)+3, 2(2-1)+1) = A\'\'(3, 3)\r\n- B\'(5, 4) → B\'\'(2(5-3)+3, 2(4-1)+1) = B\'\'(7, 7)\r\n- C\'(-2, 1) → C\'\'(2(-2-3)+3, 2(1-1)+1) = C\'\'(-7, 1)\r\n\r\n**Final Answer:**\r\n- After reflection: **A\'(3, 2), B\'(5, 4), C\'(-2, 1)**\r\n- After enlargement: **A\'\'(3, 3), B\'\'(7, 7), C\'\'(-7, 1)**\r\n\r\n![Graph of Triangle ABC under E∘R transformation](https://seekoguide.com/images/comb-of-transformation/transformation-p2.svg)\r\n\r\n### Problem 3: Reflection followed by Rotation\r\n**Question:** Find the coordinates of the image of point A(3, -5) after reflection on the line x + y = 0 followed by rotation through +90° about the origin.\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Identify the reflection line**\r\nThe line x + y = 0 can be rewritten as y = -x. This is reflection in the line y = -x.\r\n\r\n**Step 2: Apply reflection**\r\nFor reflection in line y = -x, the rule is: $P(x, y) \\rightarrow P\'(-y, -x)$\r\n- Apply to A(3, -5): $A(3, -5) \\rightarrow A\'(-(-5), -3) = A\'(5, -3)$\r\n\r\n**Step 3: Apply rotation**\r\nFor +90° rotation, the rule is: $P(x, y) \\rightarrow P\'(-y, x)$\r\n- Apply to A\'(5, -3): $A\'(5, -3) \\rightarrow A\'\'(-(-3), 5) = A\'\'(3, 5)$\r\n\r\n**Final Answer:**\r\n- Intermediate image after reflection: **A\'(5, -3)**\r\n- Final image after rotation: **A\'\'(3, 5)**\r\n\r\n### Problem 4: Triangle Reflection and Enlargement\r\n**Question:** Triangle PQR having vertices P(2, 1), Q(5, 3) and R(7, -1) is reflected on the y-axis. The image so formed is enlarged by E[(0, 0), 2]. Write the co-ordinates of the vertices images thus obtained and present the △PQR and its image in same graph paper.\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply reflection on y-axis**\r\nReflection rule: $P(x, y) \\rightarrow P\'(-x, y)$\r\n- P(2, 1) → P\'(-2, 1)\r\n- Q(5, 3) → Q\'(-5, 3)\r\n- R(7, -1) → R\'(-7, -1)\r\n\r\n**Step 2: Apply enlargement**\r\nEnlargement rule: $P(x, y) \\rightarrow P\'\'(2x, 2y)$\r\n- P\'(-2, 1) → P\'\'(-4, 2)\r\n- Q\'(-5, 3) → Q\'\'(-10, 6)\r\n- R\'(-7, -1) → R\'\'(-14, -2)\r\n\r\n**Final Answer:**\r\n- After reflection: **P\'(-2, 1), Q\'(-5, 3), R\'(-7, -1)**\r\n- After enlargement: **P\'\'(-4, 2), Q\'\'(-10, 6), R\'\'(-14, -2)**\r\n\r\n![Graph of Triangle PQR and its transformations](https://seekoguide.com/images/comb-of-transformation/22.svg)\r\n\r\n### Problem 5: Parallelogram Rotation and Enlargement\r\n**Question:** L(2, 2), O(6, 2), V(7,4) and E(3, 4) are the vertices of a parallelogram LOVE. Find the coordinates of the vertices of the images of ▱LOVE under the rotation of positive 90° about origin followed by enlargement E[(0,0), 3]. Represent the object and image on the same graph paper.\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply +90° rotation**\r\nRotation rule: $P(x, y) \\rightarrow P\'(-y, x)$\r\n- L(2, 2) → L\'(-2, 2)\r\n- O(6, 2) → O\'(-2, 6)\r\n- V(7, 4) → V\'(-4, 7)\r\n- E(3, 4) → E\'(-4, 3)\r\n\r\n**Step 2: Apply enlargement**\r\nEnlargement rule: $P(x, y) \\rightarrow P\'\'(3x, 3y)$\r\n- L\'(-2, 2) → L\'\'(-6, 6)\r\n- O\'(-2, 6) → O\'\'(-6, 18)\r\n- V\'(-4, 7) → V\'\'(-12, 21)\r\n- E\'(-4, 3) → E\'\'(-12, 9)\r\n\r\n**Final Answer:**\r\n- After rotation: **L\'(-2, 2), O\'(-2, 6), V\'(-4, 7), E\'(-4, 3)**\r\n- After enlargement: **L\'\'(-6, 6), O\'\'(-6, 18), V\'\'(-12, 21), E\'\'(-12, 9)**\r\n\r\n![Graph of Parallelogram LOVE and its transformations](https://seekoguide.com/images/comb-of-transformation/23.svg)\r\n\r\n### Problem 6: Triangle with Reflection in Line y = x\r\n**Question:** P(1, 2), Q(2, 1) and R(4, 3) are the vertices of the triangle PQR. Find the coordinates of the images of the triangle PQR under the reflection line y = x followed by the enlargement E[(0, 0), 2]. Present the △PQR and its images on the same graph.\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply reflection in y = x**\r\nReflection rule: $P(x, y) \\rightarrow P\'(y, x)$\r\n- P(1, 2) → P\'(2, 1)\r\n- Q(2, 1) → Q\'(1, 2)\r\n- R(4, 3) → R\'(3, 4)\r\n\r\n**Step 2: Apply enlargement**\r\nEnlargement rule: $P(x, y) \\rightarrow P\'\'(2x, 2y)$\r\n- P\'(2, 1) → P\'\'(4, 2)\r\n- Q\'(1, 2) → Q\'\'(2, 4)\r\n- R\'(3, 4) → R\'\'(6, 8)\r\n\r\n**Final Answer:**\r\n- After reflection: **P\'(2, 1), Q\'(1, 2), R\'(3, 4)**\r\n- After enlargement: **P\'\'(4, 2), Q\'\'(2, 4), R\'\'(6, 8)**\r\n\r\n![Graph of Triangle PQR and its transformations](https://seekoguide.com/images/comb-of-transformation/24.svg)\r\n\r\n### Problem 7: Translation followed by Enlargement\r\n**Question:** The vertices of △ABC are A(2, 0), B(3, 1) and C(1, 1). △ABC is translated by vector $\\begin{pmatrix} 2 \\\\ -3 \\end{pmatrix}$. The image so formed is enlarged by E[(0, 0), 2]. Write the coordinates of the image thus obtained and represent the object and the image on the same graph paper.\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply translation**\r\nTranslation rule: $P(x, y) \\rightarrow P\'(x+2, y-3)$\r\n- A(2, 0) → A\'(4, -3)\r\n- B(3, 1) → B\'(5, -2)\r\n- C(1, 1) → C\'(3, -2)\r\n\r\n**Step 2: Apply enlargement**\r\nEnlargement rule: $P(x, y) \\rightarrow P\'\'(2x, 2y)$\r\n- A\'(4, -3) → A\'\'(8, -6)\r\n- B\'(5, -2) → B\'\'(10, -4)\r\n- C\'(3, -2) → C\'\'(6, -4)\r\n\r\n**Final Answer:**\r\n- After translation: **A\'(4, -3), B\'(5, -2), C\'(3, -2)**\r\n- After enlargement: **A\'\'(8, -6), B\'\'(10, -4), C\'\'(6, -4)**\r\n\r\n![Graph of Triangle ABC and its transformations](https://seekoguide.com/images/comb-of-transformation/25.svg)\r\n\r\n### Problem 8: Rotation and Enlargement of Triangle CAT\r\n**Question:** C(2, 5), A(-1, 3) and T(4, 1) are the vertices Triangle CAT. Find the coordinates of the vertices of the image of △CAT under the rotation of positive 90° about origin followed by enlargement E[(0,0), 2]. Represent the object and images on the same graph paper.\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply +90° rotation**\r\nRotation rule: $P(x, y) \\rightarrow P\'(-y, x)$\r\n- C(2, 5) → C\'(-5, 2)\r\n- A(-1, 3) → A\'(-3, -1)\r\n- T(4, 1) → T\'(-1, 4)\r\n\r\n**Step 2: Apply enlargement**\r\nEnlargement rule: $P(x, y) \\rightarrow P\'\'(2x, 2y)$\r\n- C\'(-5, 2) → C\'\'(-10, 4)\r\n- A\'(-3, -1) → A\'\'(-6, -2)\r\n- T\'(-1, 4) → T\'\'(-2, 8)\r\n\r\n**Final Answer:**\r\n- After rotation: **C\'(-5, 2), A\'(-3, -1), T\'(-1, 4)**\r\n- After enlargement: **C\'\'(-10, 4), A\'\'(-6, -2), T\'\'(-2, 8)**\r\n\r\n![Graph of Triangle CAT and its transformations](https://seekoguide.com/images/comb-of-transformation/26.svg)\r\n\r\n---\r\n\r\n## 🧮 Practice Problems with Solutions\r\n\r\n### Practice Set A: Equivalent Single Transformations\r\n\r\n**Problem A1:** State the single transformation equivalent to the combination of reflections on the X-axis and Y-axis respectively. Using this single transformation find the coordinates of the vertices of the image of △PQR having vertices P(4, 3), Q(1, 1) and R(5, -1). Also draw the object and image on the same graph.\r\n\r\n**Solution:**\r\n**Step 1: Find the equivalent transformation**\r\n- First reflection on x-axis: P(x, y) → P\'(x, -y)\r\n- Second reflection on y-axis: P\'(x, -y) → P\'\'(-x, -y)\r\n- Combined transformation: P(x, y) → P\'\'(-x, -y)\r\n- This is equivalent to **rotation of ±180° about the origin**\r\n\r\n**Step 2: Apply to triangle vertices**\r\n- P(4, 3) → P\'(-4, -3)\r\n- Q(1, 1) → Q\'(-1, -1)\r\n- R(5, -1) → R\'(-5, 1)\r\n\r\n![Graph showing triangle PQR and its image](https://seekoguide.com/images/comb-of-transformation/27.svg)\r\n\r\n**Answer: Single transformation = 180° rotation; Final vertices: P\'(-4, -3), Q\'(-1, -1), R\'(-5, 1)**\r\n\r\n**Problem A2:** A △ABC with vertices A(1, 2), B(4, -1) and C(2, 5) is reflected successively in the line y = x and the X-axis. Find the co-ordinates and graphically represent the image under these transformations. State also the single transformation given by the combinations of these transformations.\r\n\r\n**Solution:**\r\n**Step 1: Reflection in y = x**\r\nRule: P(x, y) → P\'(y, x)\r\n- A(1, 2) → A\'(2, 1)\r\n- B(4, -1) → B\'(-1, 4)\r\n- C(2, 5) → C\'(5, 2)\r\n\r\n**Step 2: Reflection in x-axis**\r\nRule: P(x, y) → P\'\'(x, -y)\r\n- A\'(2, 1) → A\'\'(2, -1)\r\n- B\'(-1, 4) → B\'\'(-1, -4)\r\n- C\'(5, 2) → C\'\'(5, -2)\r\n\r\n**Step 3: Find single equivalent transformation**\r\nCombined rule: P(x, y) → P\'\'(y, -x)\r\nThis is equivalent to **rotation of -90° about the origin**\r\n\r\n![Graph showing triangle ABC and its transformations](https://seekoguide.com/images/comb-of-transformation/29.svg)\r\n\r\n**Answer: Final vertices: A\'\'(2, -1), B\'\'(-1, -4), C\'\'(5, -2); Single transformation = -90° rotation**\r\n\r\n### Practice Set B: Complex Combined Transformations\r\n\r\n**Problem B1:** Translate the △PQR with vertices P(2, 4), Q(-1, 2) and R(5, -1) by translation vector T = $\\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}$ and then reflect the image so obtained in the line x + y = 0. Write the coordinates of the vertices of the image and draw the object and images on the same graph paper.\r\n\r\n**Solution:**\r\n**Step 1: Apply translation**\r\nTranslation rule: P(x, y) → P\'(x-2, y+3)\r\n- P(2, 4) → P\'(0, 7)\r\n- Q(-1, 2) → Q\'(-3, 5)\r\n- R(5, -1) → R\'(3, 2)\r\n\r\n**Step 2: Apply reflection in x + y = 0**\r\nFor line x + y = 0 (y = -x), rule: P(x, y) → P\'\'(-y, -x)\r\n- P\'(0, 7) → P\'\'(-7, 0)\r\n- Q\'(-3, 5) → Q\'\'(-5, 3)\r\n- R\'(3, 2) → R\'\'(-2, -3)\r\n\r\n![Graph showing triangle PQR and its transformations](https://seekoguide.com/images/comb-of-transformation/32.svg)\r\n\r\n**Answer: Final vertices: P\'\'(-7, 0), Q\'\'(-5, 3), R\'\'(-2, -3)**\r\n\r\n**Problem B2:** Draw the image in graph that is formed when a triangle with vertices A(1, 0), B(2, 1) and C(3, -1) is translated by $\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix}$. Reflect the image in the line x = 2 and also draw the image in the same graph.\r\n\r\n**Solution:**\r\n**Step 1: Apply translation**\r\nTranslation rule: P(x, y) → P\'(x+1, y+2)\r\n- A(1, 0) → A\'(2, 2)\r\n- B(2, 1) → B\'(3, 3)\r\n- C(3, -1) → C\'(4, 1)\r\n\r\n**Step 2: Apply reflection in x = 2**\r\nRule: P(x, y) → P\'\'(4-x, y)\r\n- A\'(2, 2) → A\'\'(2, 2)\r\n- B\'(3, 3) → B\'\'(1, 3)\r\n- C\'(4, 1) → C\'\'(0, 1)\r\n\r\n![Graph showing triangle ABC and its transformations](https://seekoguide.com/images/comb-of-transformation/33.svg)\r\n\r\n**Answer: Final vertices: A\'\'(2, 2), B\'\'(1, 3), C\'\'(0, 1)**\r\n\r\n### Practice Set C: Advanced Reflection Problems\r\n\r\n**Problem C1:** A triangle with vertices N(1, 2), V(4, -1) and M(2, 5) is reflected successively in the lines x = -1 and y = 2. Find by stating coordinates and represent the images graphically under these transformations. Also state the single transformation given by these transformations.\r\n\r\n**Solution:**\r\n**Step 1: Reflection in x = -1**\r\nRule: P(x, y) → P\'(-2-x, y)\r\n- N(1, 2) → N\'(-3, 2)\r\n- V(4, -1) → V\'(-6, -1)\r\n- M(2, 5) → M\'(-4, 5)\r\n\r\n**Step 2: Reflection in y = 2**\r\nRule: P(x, y) → P\'\'(x, 4-y)\r\n- N\'(-3, 2) → N\'\'(-3, 2)\r\n- V\'(-6, -1) → V\'\'(-6, 5)\r\n- M\'(-4, 5) → M\'\'(-4, -1)\r\n\r\n**Step 3: Find single transformation**\r\nThe combined effect is reflection in line x = -1 (since reflecting in two perpendicular lines gives reflection in the line equidistant from them).\r\n\r\n![Graph showing triangle NVM and its transformations](https://seekoguide.com/images/comb-of-transformation/34.svg)\r\n\r\n**Answer: Final vertices: N\'\'(-3, 2), V\'\'(-6, 5), M\'\'(-4, -1); Single transformation = reflection in x = -1**\r\n\r\n### Practice Set D: Rotation and Enlargement Problems\r\n\r\n**Problem D1:** The vertices of △ABC are A(2, 5), B(-2, 3) and C(4, 1). △ABC is rotated through negative quarter turn about the origin and the image so obtained is enlarged by taking origin as the centre and scale factor 2. Hence, find the co-ordinates of the image of △ABC and show the object and image on the same graph.\r\n\r\n**Solution:**\r\n**Step 1: Apply -90° rotation (negative quarter turn)**\r\nRule: P(x, y) → P\'(y, -x)\r\n- A(2, 5) → A\'(5, -2)\r\n- B(-2, 3) → B\'(3, 2)\r\n- C(4, 1) → C\'(1, -4)\r\n\r\n**Step 2: Apply enlargement from origin with scale factor 2**\r\nRule: P(x, y) → P\'\'(2x, 2y)\r\n- A\'(5, -2) → A\'\'(10, -4)\r\n- B\'(3, 2) → B\'\'(6, 4)\r\n- C\'(1, -4) → C\'\'(2, -8)\r\n\r\n![Graph showing triangle ABC and its transformations](https://seekoguide.com/images/comb-of-transformation/30.svg)\r\n\r\n**Answer: Final vertices: A\'\'(10, -4), B\'\'(6, 4), C\'\'(2, -8)**\r\n\r\n**Problem D2:** K(2, 5), L(-1, 3) and M(4, 1) are the vertices of a triangle KLM. Find the co-ordinates of the vertices of the images of △KLM under the rotation of negative 90° about the origin followed by the enlargement E[(0, 0), 2]. Present object and its images on the same graph paper.\r\n\r\n**Solution:**\r\n**Step 1: Apply -90° rotation**\r\nRule: P(x, y) → P\'(y, -x)\r\n- K(2, 5) → K\'(5, -2)\r\n- L(-1, 3) → L\'(3, 1)\r\n- M(4, 1) → M\'(1, -4)\r\n\r\n**Step 2: Apply enlargement**\r\nRule: P(x, y) → P\'\'(2x, 2y)\r\n- K\'(5, -2) → K\'\'(10, -4)\r\n- L\'(3, 1) → L\'\'(6, 2)\r\n- M\'(1, -4) → M\'\'(2, -8)\r\n\r\n![Graph showing triangle KLM and its transformations](https://seekoguide.com/images/comb-of-transformation/31.svg)\r\n\r\n**Answer: Final vertices: K\'\'(10, -4), L\'\'(6, 2), M\'\'(2, -8)**\r\n\r\n---\r\n\r\n## 🎨 Understanding Order of Operations\r\n\r\n### Why Order Matters\r\n\r\nThe order in which transformations are applied is **crucial**. Different orders can produce completely different results.\r\n\r\n**Example Demonstration:**\r\nConsider point P(2, 1) with two transformations:\r\n- T₁: Reflection in x-axis\r\n- T₂: Translation by $\\begin{pmatrix} 3 \\\\ 0 \\end{pmatrix}$\r\n\r\n**Order 1: T₁ then T₂**\r\n- P(2, 1) → P\'(2, -1) → P\'\'(5, -1)\r\n\r\n**Order 2: T₂ then T₁**\r\n- P(2, 1) → P\'(5, 1) → P\'\'(5, -1)\r\n\r\n**Result:** Same final point, but this is not always the case!\r\n\r\n**Counter-example with different final results:**\r\nPoint P(1, 2) with transformations:\r\n- T₁: +90° rotation\r\n- T₂: Translation by $\\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix}$\r\n\r\n**Order 1: T₁ then T₂**\r\n- P(1, 2) → (-2, 1) → (-1, 1)\r\n\r\n**Order 2: T₂ then T₁**\r\n- P(1, 2) → (2, 2) → (-2, 2)\r\n\r\n**Different results!** Order matters significantly.\r\n\r\n---\r\n\r\n## 💡 Problem-Solving Strategies\r\n\r\n### Strategy 1: Step-by-Step Method\r\n1. **Identify** all transformations and their order\r\n2. **Apply** each transformation sequentially\r\n3. **Check** your work by verifying intermediate steps\r\n4. **State** both intermediate and final results clearly\r\n\r\n### Strategy 2: Combined Rule Method\r\n1. **Derive** the single combined transformation rule\r\n2. **Apply** this rule directly to find the final image\r\n3. **Verify** by checking with step-by-step method\r\n\r\n### Strategy 3: Working Backwards\r\n1. **Start** with the final image\r\n2. **Reverse** each transformation in opposite order\r\n3. **Apply** inverse transformations to find original point\r\n\r\n---\r\n\r\n## 📊 Common Mistakes to Avoid\r\n\r\n### Mistake 1: Wrong Order of Operations\r\n❌ **Wrong:** Applying transformations in written order when composition notation indicates reverse order\r\n✅ **Correct:** For f ∘ g, apply g first, then f\r\n\r\n### Mistake 2: Sign Errors in Rotations\r\n❌ **Wrong:** Confusing +90° and -90° rotation rules\r\n✅ **Correct:** \r\n- +90°: (x, y) → (-y, x)\r\n- -90°: (x, y) → (y, -x)\r\n\r\n### Mistake 3: Reflection Line Confusion\r\n❌ **Wrong:** Using wrong reflection formulas\r\n✅ **Correct:** Always identify the reflection line clearly before applying rules\r\n\r\n---\r\n\r\n## 🎯 Examination Tips\r\n\r\n### Time-Saving Techniques\r\n1. **Master the basic rules** - know them by heart\r\n2. **Use systematic notation** - label each intermediate step\r\n3. **Double-check signs** - most errors occur with positive/negative signs\r\n4. **Verify with graphical thinking** - visualize transformations mentally\r\n\r\n### Common Exam Question Types\r\n1. **Direct application**: Apply given transformations in sequence\r\n2. **Finding unknowns**: Given final result, find original point or unknown parameters\r\n3. **Equivalent transformations**: Find single transformation equivalent to combination\r\n4. **Order dependency**: Compare results of different transformation orders\r\n\r\n---\r\n\r\n## 🏆 Advanced Applications\r\n\r\n### Composite Transformations in Real World\r\n- **Computer Graphics**: Rotating and scaling 3D models\r\n- **Engineering**: Coordinate system transformations\r\n- **Physics**: Reference frame changes\r\n- **Art and Design**: Creating patterns and symmetries\r\n\r\n### Extension Topics\r\n- Transformations in 3D space\r\n- Matrix representations of transformations\r\n- Inverse transformations\r\n- Transformation groups in advanced mathematics\r\n\r\n---\r\n\r\n## 📝 Summary Checklist\r\n\r\nBy the end of this guide, you should be able to:\r\n\r\n✅ **Apply basic transformations** (reflection, rotation, translation, enlargement) to points and shapes  \r\n✅ **Combine multiple transformations** in the correct order using proper notation  \r\n✅ **Work with triangles, parallelograms, and other polygons** in transformation problems  \r\n✅ **Handle reflection in various lines** (x-axis, y-axis, y=x, y=-x, x=k, y=k)  \r\n✅ **Execute rotations** (+90°, -90°, 180°) and understand quarter-turns  \r\n✅ **Apply translations** using vector notation and coordinate shifts  \r\n✅ **Perform enlargements** from origin and from other centers with various scale factors  \r\n✅ **Interpret composition notation** (E∘R, r₁∘r₂) correctly  \r\n✅ **Find equivalent single transformations** for combined operations  \r\n✅ **Understand why order matters** in combined transformations  \r\n✅ **Solve for unknown coordinates and parameters** in complex problems  \r\n✅ **Use efficient problem-solving strategies** for multi-step transformations  \r\n✅ **Avoid common mistakes** with signs, order, and transformation rules  \r\n✅ **Work with graphical representations** and coordinate plotting  \r\n✅ **Handle negative quarter-turns** and clockwise/counterclockwise rotations  \r\n✅ **Solve real examination-style problems** with complete step-by-step solutions  \r\n\r\n---\r\n\r\n## 🎓 Final Words\r\n\r\nMastering combined transformations requires **practice, patience, and systematic thinking**. The key to success is:\r\n\r\n1. **Understanding each basic transformation thoroughly**\r\n2. **Practicing the step-by-step approach consistently**\r\n3. **Checking your work at each stage**\r\n4. **Learning from mistakes and understanding why they occur**\r\n\r\nRemember: **Every complex transformation is just a series of simple steps applied correctly in the right order.**\r\n\r\nKeep practicing, stay organized in your approach, and soon you\'ll master even the most complex combined transformation problems! 🚀\r\n\r\n---\r\n\r\n', 1, '2025-07-25 01:11:38', '2025-09-18 14:27:45'),
(155, 14245, '# Circle Inversion: Complete Study Guide for Geometric Inversion\r\n## A Comprehensive Guide to Inversion with Respect to Circles\r\n\r\n---\r\n\r\n## 🎯 Introduction: Understanding Circle Inversion\r\n\r\nWelcome to the complete guide on **Circle Inversion** (also known as **Geometric Inversion**). Circle inversion is a powerful transformation in geometry that maps points inside a circle to points outside the circle and vice versa, while preserving certain geometric properties.\r\n\r\nThis transformation has fascinating applications in:\r\n- Complex analysis and conformal mapping\r\n- Geometric constructions and proofs\r\n- Non-Euclidean geometry\r\n- Physics and engineering problems\r\n- Advanced mathematics competitions\r\n\r\n**Key Learning Objective**: Master the concept of inversion, understand the fundamental inversion formula, and solve complex problems involving inverse points with respect to circles. 📐\r\n\r\n---\r\n\r\n## 🔑 Fundamental Concepts of Circle Inversion\r\n\r\n### What is Circle Inversion?\r\n\r\n**Circle Inversion** is a transformation that maps each point P in the plane (except the center) to a corresponding point P\' such that:\r\n- The center O, point P, and point P\' are **collinear**\r\n- The relationship **OP × OP\' = r²** holds (where r is the radius of the circle of inversion)\r\n\r\n### Basic Properties of Inversion\r\n\r\n1. **Reciprocal Relationship**: If P\' is the inverse of P, then P is also the inverse of P\'\r\n2. **Points on the Circle**: Points on the circle of inversion map to themselves\r\n3. **Inside ↔ Outside**: Points inside the circle map to points outside, and vice versa\r\n4. **Center Behavior**: The center of inversion has no inverse (maps to infinity)\r\n\r\n---\r\n\r\n## 📐 Essential Inversion Formulas\r\n\r\n### 1. For Circle Centered at Origin\r\n\r\n**Circle of Inversion**: $x^2 + y^2 = r^2$\r\n\r\n**Inversion Formula**:\r\n$$P\'(x\', y\') = \\left(\\frac{r^2 x}{x^2 + y^2}, \\frac{r^2 y}{x^2 + y^2}\\right)$$\r\n\r\n**Key Properties**:\r\n- Distance relationship: $|OP| \\times |OP\'| = r^2$\r\n- If $|OP| = d$, then $|OP\'| = \\frac{r^2}{d}$\r\n\r\n### 2. For Circle Centered at Point (h, k)\r\n\r\n**Circle of Inversion**: $(x - h)^2 + (y - k)^2 = r^2$\r\n\r\n**Inversion Formula**:\r\n$$P\'(x\', y\') = \\left(\\frac{r^2(x - h)}{(x - h)^2 + (y - k)^2} + h, \\frac{r^2(y - k)}{(x - h)^2 + (y - k)^2} + k\\right)$$\r\n\r\n**Key Properties**:\r\n- The center (h, k) is the center of inversion\r\n- Distance relationship: $|CP| \\times |CP\'| = r^2$ where C is the center\r\n\r\n---\r\n\r\n## 📚 Worked Examples and Solutions\r\n\r\n### Problem 1: Understanding the Basic Inversion Relationship\r\n**Question:** In the figure, O is centre of circle, OP = 4 units, OQ = 8 units and OP\' = 16 units, write the relation between P and P\'.\r\n\r\n![Circle inversion diagram](https://seekoguide.com/images/inversion/1.svg)\r\n\r\n**Complete Solution:**\r\n\r\n**Given Information:**\r\n- OP = 4 units\r\n- OQ = 8 units (radius of circle)\r\n- OP\' = 16 units\r\n\r\n**Step 1: Check the inversion relationship**\r\nFor inversion with respect to a circle, we must have: $OP \\times OP\' = r^2$\r\n\r\n**Step 2: Verify the relationship**\r\n- $OP \\times OP\' = 4 \\times 16 = 64$\r\n- $r^2 = OQ^2 = 8^2 = 64$\r\n- Since $64 = 64$, the relationship holds!\r\n\r\n**Step 3: State the conclusion**\r\nThe relationship is: $OP \\times OP\' = r^2$\r\n\r\n**Final Answer:** P\' is the **inversion point of P** (and vice versa) with respect to the circle centered at O with radius 8 units.\r\n\r\n### Problem 2: Finding Inverse with Respect to Circle Centered at Origin\r\n**Question:** Find the inverse image of the point (4, 5) with respect to the circle $x^2 + y^2 = 10$.\r\n\r\n**Complete Solution:**\r\n\r\n**Given Information:**\r\n- Circle: $x^2 + y^2 = 10$\r\n- Center of circle: (0, 0)\r\n- Radius: $r^2 = 10$ (so $r = \\sqrt{10}$)\r\n- Object point P(x, y) = (4, 5)\r\n- Inversion point P\'(x\', y\') = ?\r\n\r\n**Step 1: Apply the inversion formula**\r\nFor circle centered at origin:\r\n$$P\'(x\', y\') = \\left(\\frac{r^2 x}{x^2 + y^2}, \\frac{r^2 y}{x^2 + y^2}\\right)$$\r\n\r\n**Step 2: Substitute the values**\r\n$$P\'(x\', y\') = \\left(\\frac{10 \\times 4}{4^2 + 5^2}, \\frac{10 \\times 5}{4^2 + 5^2}\\right)$$\r\n\r\n**Step 3: Calculate the denominator**\r\n$4^2 + 5^2 = 16 + 25 = 41$\r\n\r\n**Step 4: Complete the calculation**\r\n$$P\'(x\', y\') = \\left(\\frac{40}{41}, \\frac{50}{41}\\right)$$\r\n\r\n**Final Answer:** The inverse of point (4, 5) is $\\left(\\frac{40}{41}, \\frac{50}{41}\\right)$\r\n\r\n**Verification:** $OP \\times OP\' = \\sqrt{41} \\times \\frac{\\sqrt{40^2 + 50^2}}{41} = \\sqrt{41} \\times \\frac{\\sqrt{4100}}{41} = \\sqrt{41} \\times \\frac{\\sqrt{41 \\times 100}}{41} = 10$ ✓\r\n\r\n### Problem 3: Finding Inverse with Respect to Circle with Different Center\r\n**Question:** Find the inverse image of the point (3, 4) about the circle $(x - 2)^2 + (y - 2)^2 = 36$.\r\n\r\n**Complete Solution:**\r\n\r\n**Given Information:**\r\n- Circle: $(x - 2)^2 + (y - 2)^2 = 36$\r\n- Center of circle (h, k) = (2, 2)\r\n- Radius: r = 6 (since $r^2 = 36$)\r\n- Object point (x, y) = (3, 4)\r\n- Inversion point (x\', y\') = ?\r\n\r\n**Step 1: Apply the inversion formula for non-origin center**\r\n$$P\'(x\', y\') = \\left(\\frac{r^2(x - h)}{(x - h)^2 + (y - k)^2} + h, \\frac{r^2(y - k)}{(x - h)^2 + (y - k)^2} + k\\right)$$\r\n\r\n**Step 2: Calculate the distance from point to center**\r\n- $(x - h) = 3 - 2 = 1$\r\n- $(y - k) = 4 - 2 = 2$\r\n- $(x - h)^2 + (y - k)^2 = 1^2 + 2^2 = 1 + 4 = 5$\r\n\r\n**Step 3: Substitute into the formula**\r\n$$P\'(x\', y\') = \\left(\\frac{36 \\times 1}{5} + 2, \\frac{36 \\times 2}{5} + 2\\right)$$\r\n\r\n**Step 4: Simplify**\r\n$$P\'(x\', y\') = \\left(\\frac{36}{5} + 2, \\frac{72}{5} + 2\\right)$$\r\n$$P\'(x\', y\') = \\left(\\frac{36 + 10}{5}, \\frac{72 + 10}{5}\\right)$$\r\n$$P\'(x\', y\') = \\left(\\frac{46}{5}, \\frac{82}{5}\\right)$$\r\n\r\n**Final Answer:** The inverse of point (3, 4) about the given circle is $\\left(\\frac{46}{5}, \\frac{82}{5}\\right)$\r\n\r\n---\r\n\r\n## 🧮 Practice Problems with Solutions\r\n\r\n### Practice Set A: Basic Inversion Problems\r\n\r\n**Problem A1:** Find the inverse of point (6, 8) with respect to the circle $x^2 + y^2 = 25$.\r\n\r\n**Solution:**\r\n- Center: (0, 0), $r^2 = 25$\r\n- $x^2 + y^2 = 6^2 + 8^2 = 36 + 64 = 100$\r\n- $P\'(x\', y\') = \\left(\\frac{25 \\times 6}{100}, \\frac{25 \\times 8}{100}\\right) = \\left(\\frac{3}{2}, 2\\right)$\r\n\r\n**Answer: $\\left(\\frac{3}{2}, 2\\right)$**\r\n\r\n**Problem A2:** If P(2, 3) and P\'(a, b) are inverse points with respect to circle $x^2 + y^2 = 26$, find a and b.\r\n\r\n**Solution:**\r\n- $r^2 = 26$, $x^2 + y^2 = 4 + 9 = 13$\r\n- $a = \\frac{26 \\times 2}{13} = 4$\r\n- $b = \\frac{26 \\times 3}{13} = 6$\r\n\r\n**Answer: a = 4, b = 6**\r\n\r\n### Practice Set B: Advanced Inversion Problems\r\n\r\n**Problem B1:** Find the inverse of point (5, 1) with respect to circle $(x - 3)^2 + (y - 2)^2 = 16$.\r\n\r\n**Solution:**\r\n- Center: (3, 2), $r^2 = 16$\r\n- $(x - h) = 5 - 3 = 2$, $(y - k) = 1 - 2 = -1$\r\n- $(x - h)^2 + (y - k)^2 = 4 + 1 = 5$\r\n- $x\' = \\frac{16 \\times 2}{5} + 3 = \\frac{32}{5} + 3 = \\frac{47}{5}$\r\n- $y\' = \\frac{16 \\times (-1)}{5} + 2 = -\\frac{16}{5} + 2 = -\\frac{6}{5}$\r\n\r\n**Answer: $\\left(\\frac{47}{5}, -\\frac{6}{5}\\right)$**\r\n\r\n**Problem B2:** Show that if two points are inverse with respect to a circle, then the circle is the perpendicular bisector of the line segment joining the two points when extended.\r\n\r\n**Solution:**\r\nThis is a theorem in inversion geometry. The proof involves showing that any circle orthogonal to the circle of inversion passes through both P and P\', making the inversion circle the locus of centers of all such circles.\r\n\r\n### Practice Set C: Applications and Properties\r\n\r\n**Problem C1:** Find the equation of the circle that passes through point (1, 2) and its inverse with respect to circle $x^2 + y^2 = 9$.\r\n\r\n**Solution:**\r\n**Step 1:** Find inverse of (1, 2)\r\n- $P\' = \\left(\\frac{9 \\times 1}{5}, \\frac{9 \\times 2}{5}\\right) = \\left(\\frac{9}{5}, \\frac{18}{5}\\right)$\r\n\r\n**Step 2:** Any circle passing through both points and orthogonal to $x^2 + y^2 = 9$\r\n- The required circle has the form: $x^2 + y^2 + Dx + Ey + F = 0$ where it\'s orthogonal to the inversion circle.\r\n\r\n**Problem C2:** Prove that inversion preserves angles.\r\n\r\n**Solution:**\r\nThis is a fundamental theorem: **Inversion is a conformal transformation**, meaning it preserves angles between curves at their intersection points.\r\n\r\n---\r\n\r\n## 🎨 Special Properties of Inversion\r\n\r\n### Geometric Properties Preserved Under Inversion\r\n\r\n1. **Angle Preservation**: Angles between curves are preserved\r\n2. **Circle/Line Mapping**: \r\n   - Circles through the center → Lines not through the center\r\n   - Circles not through the center → Circles not through the center\r\n   - Lines through the center → Lines through the center\r\n   - Lines not through the center → Circles through the center\r\n\r\n### Important Theorems\r\n\r\n**Theorem 1**: The inverse of a line passing through the center of inversion is the line itself.\r\n\r\n**Theorem 2**: The inverse of a line not passing through the center of inversion is a circle passing through the center of inversion.\r\n\r\n**Theorem 3**: The inverse of a circle passing through the center of inversion is a line not passing through the center.\r\n\r\n**Theorem 4**: The inverse of a circle not passing through the center of inversion is another circle not passing through the center.\r\n\r\n---\r\n\r\n## 💡 Problem-Solving Strategies\r\n\r\n### Strategy 1: Direct Formula Application\r\n1. **Identify** the circle of inversion (center and radius)\r\n2. **Determine** which formula to use (origin-centered or general)\r\n3. **Substitute** coordinates carefully\r\n4. **Simplify** step by step\r\n5. **Verify** using the fundamental relationship\r\n\r\n### Strategy 2: Geometric Approach\r\n1. **Draw** the configuration\r\n2. **Use** the property $OP \\times OP\' = r^2$\r\n3. **Apply** similar triangles when appropriate\r\n4. **Check** collinearity of O, P, and P\'\r\n\r\n### Strategy 3: Verification Method\r\n1. **Calculate** distances OP and OP\'\r\n2. **Verify** that $OP \\times OP\' = r^2$\r\n3. **Check** that O, P, P\' are collinear\r\n\r\n---\r\n\r\n## 📊 Common Mistakes to Avoid\r\n\r\n### Mistake 1: Wrong Formula Selection\r\n❌ **Wrong:** Using origin formula for non-origin centered circles\r\n✅ **Correct:** Always check the center of the circle of inversion\r\n\r\n### Mistake 2: Arithmetic Errors\r\n❌ **Wrong:** Making calculation errors in fractions\r\n✅ **Correct:** Double-check all arithmetic, especially with fractions\r\n\r\n### Mistake 3: Sign Confusion\r\n❌ **Wrong:** Missing negative signs in coordinates\r\n✅ **Correct:** Pay careful attention to coordinate signs\r\n\r\n### Mistake 4: Distance Calculation\r\n❌ **Wrong:** Using wrong distance in non-origin cases\r\n✅ **Correct:** Always calculate distance from the center of inversion\r\n\r\n---\r\n\r\n## 🎯 Examination Tips\r\n\r\n### Time-Saving Techniques\r\n1. **Master both formulas** - origin and general center cases\r\n2. **Practice fraction arithmetic** - most answers involve fractions\r\n3. **Use verification** - check your answer with $OP \\times OP\' = r^2$\r\n4. **Draw diagrams** when possible for geometric insight\r\n\r\n### Common Exam Question Types\r\n1. **Direct inversion**: Find inverse of given point\r\n2. **Finding unknowns**: Given one point and its inverse, find missing coordinates\r\n3. **Relationship verification**: Prove two points are inverse\r\n4. **Applications**: Using inversion in geometric constructions\r\n\r\n---\r\n\r\n## 🏆 Advanced Applications\r\n\r\n### Applications in Mathematics\r\n- **Complex Analysis**: Möbius transformations\r\n- **Hyperbolic Geometry**: Poincaré disk model\r\n- **Geometric Proofs**: Elegant solutions using inversion\r\n- **Conformal Mapping**: Preserving angles in transformations\r\n\r\n### Applications in Physics and Engineering\r\n- **Electrostatics**: Method of images\r\n- **Fluid Dynamics**: Flow around obstacles\r\n- **Optics**: Lens design and ray tracing\r\n- **Antenna Theory**: Radiation patterns\r\n\r\n---\r\n\r\n## 📝 Summary Checklist\r\n\r\nBy the end of this guide, you should be able to:\r\n\r\n✅ **Understand the concept** of circle inversion and its geometric meaning  \r\n✅ **Apply inversion formulas** for both origin-centered and general circles  \r\n✅ **Find inverse points** given any point and circle of inversion  \r\n✅ **Verify inversion relationships** using the fundamental property $OP \\times OP\' = r^2$  \r\n✅ **Work with fractional coordinates** accurately in inversion problems  \r\n✅ **Understand geometric properties** preserved under inversion  \r\n✅ **Solve complex inversion problems** involving multiple steps  \r\n✅ **Apply inversion** to geometric constructions and proofs  \r\n✅ **Recognize applications** of inversion in advanced mathematics  \r\n✅ **Use verification methods** to check solutions  \r\n\r\n---\r\n\r\n## 🎓 Final Words\r\n\r\nCircle inversion is one of the most elegant and powerful transformations in geometry. It provides:\r\n\r\n1. **Beautiful geometric insights** into the nature of circles and lines\r\n2. **Powerful problem-solving tools** for complex geometric constructions\r\n3. **Deep connections** to advanced areas of mathematics\r\n4. **Practical applications** in physics and engineering\r\n\r\n**Key to Success:**\r\n- **Practice the formulas** until they become second nature\r\n- **Understand the geometric meaning** behind the algebraic calculations  \r\n- **Visualize the transformation** - points inside map outside and vice versa\r\n- **Use verification** as a powerful check on your work\r\n\r\nRemember: **Inversion turns the inside out and the outside in, but angles stay the same!** \r\n\r\nThis transformation has fascinated mathematicians for centuries and continues to provide elegant solutions to seemingly difficult problems. Master it, and you\'ll have a powerful tool in your mathematical toolkit! 🚀\r\n\r\n---\r\n\r\n*This guide provides comprehensive coverage of circle inversion with detailed worked examples, practice problems, and advanced applications. Use it as your complete reference for mastering geometric inversion.*\r\n', 1, '2025-07-25 01:11:38', '2025-09-18 14:27:45');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(156, 13121, '# Matrix Transformations: Complete Study Guide for Linear Transformations\r\n\r\n## A Comprehensive Guide to Matrix Representations of Geometric Transformations\r\n\r\n  \r\n\r\n---\r\n\r\n  \r\n\r\n## 🎯 Introduction: Understanding Matrix Transformations\r\n\r\n  \r\n\r\nWelcome to the complete guide on **Matrix Transformations**. This powerful area of mathematics combines linear algebra with coordinate geometry, allowing us to represent complex geometric transformations using matrix operations.\r\n\r\n  \r\n\r\nMatrix transformations are fundamental in:\r\n\r\n- Computer graphics and animation\r\n\r\n- Engineering and physics simulations\r\n\r\n- Data analysis and machine learning\r\n\r\n- Advanced coordinate geometry\r\n\r\n- Linear algebra applications\r\n\r\n  \r\n\r\n**Key Learning Objective**: Master the matrix representations of geometric transformations, understand eigenvalues and eigenvectors, and solve complex problems involving linear transformations of points and shapes. 📐\r\n\r\n  \r\n\r\n---\r\n\r\n  \r\n\r\n## 🔑 Essential Matrix Transformation Concepts\r\n\r\n  \r\n\r\n### What is a Matrix Transformation?\r\n\r\n  \r\n\r\nA **Matrix Transformation** is a function that transforms points in the plane (or space) using matrix multiplication. For a 2×2 transformation matrix **A** and point **P**, the transformed point **P\'** is given by:\r\n\r\n  \r\n\r\n$$\\mathbf{P\'} = \\mathbf{A} \\cdot \\mathbf{P}$$\r\n\r\n  \r\n\r\n### Standard Form of 2D Transformation Matrix\r\n\r\n  \r\n\r\n$$\\mathbf{A} = \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}$$\r\n\r\n  \r\n\r\nFor point $P(x, y)$:\r\n\r\n$$\\begin{bmatrix} x\' \\\\ y\' \\end{bmatrix} = \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix} = \\begin{bmatrix} ax + by \\\\ cx + dy \\end{bmatrix}$$\r\n\r\n  \r\n\r\n---\r\n\r\n  \r\n\r\n## 📐 Standard Transformation Matrices\r\n\r\n  \r\n\r\n### 1. Rotation Matrices\r\n\r\n  \r\n\r\n**Rotation by angle θ (counterclockwise):**\r\n\r\n$$\\mathbf{R}_θ = \\begin{bmatrix} \\cos θ & -\\sin θ \\\\ \\sin θ & \\cos θ \\end{bmatrix}$$\r\n\r\n  \r\n\r\n**Special Cases:**\r\n\r\n-  **90° rotation:** $\\begin{bmatrix} 0 & -1 \\\\ 1 & 0  \\end{bmatrix}$\r\n\r\n-  **180° rotation:** $\\begin{bmatrix} -1 & 0 \\\\ 0 & -1  \\end{bmatrix}$\r\n\r\n-  **270° rotation:** $\\begin{bmatrix} 0 & 1 \\\\ -1 & 0  \\end{bmatrix}$\r\n\r\n  \r\n\r\n### 2. Reflection Matrices\r\n\r\n  \r\n\r\n**Reflection about x-axis:** $\\begin{bmatrix} 1 & 0 \\\\ 0 & -1  \\end{bmatrix}$\r\n\r\n  \r\n\r\n**Reflection about y-axis:** $\\begin{bmatrix} -1 & 0 \\\\ 0 & 1  \\end{bmatrix}$\r\n\r\n  \r\n\r\n**Reflection about line y = x:** $\\begin{bmatrix} 0 & 1 \\\\ 1 & 0  \\end{bmatrix}$\r\n\r\n  \r\n\r\n**Reflection about line y = -x:** $\\begin{bmatrix} 0 & -1 \\\\ -1 & 0  \\end{bmatrix}$\r\n\r\n  \r\n\r\n### 3. Scaling (Dilation) Matrices\r\n\r\n  \r\n\r\n**Uniform scaling by factor k:** $\\begin{bmatrix} k & 0 \\\\ 0 & k \\end{bmatrix}$\r\n\r\n  \r\n\r\n**Non-uniform scaling:** $\\begin{bmatrix} k_x & 0 \\\\ 0 & k_y \\end{bmatrix}$\r\n\r\n  \r\n\r\n### 4. Shearing Matrices\r\n\r\n  \r\n\r\n**Horizontal shear:** $\\begin{bmatrix} 1 & k \\\\ 0 & 1  \\end{bmatrix}$\r\n\r\n  \r\n\r\n**Vertical shear:** $\\begin{bmatrix} 1 & 0 \\\\ k & 1  \\end{bmatrix}$\r\n\r\n  \r\n\r\n---\r\n\r\n  \r\n\r\n## 📚 Worked Examples and Solutions\r\n\r\n  \r\n  \r\n\r\n### Problem 1: Identifying Transformation Types (From Raw Data)\r\n\r\n**Question:** Which transformation does the matrix $\\begin{bmatrix} 0 & -1 \\\\ -1 & 0  \\end{bmatrix}$ represent?\r\n\r\n  \r\n\r\n**Complete Solution:**\r\n\r\n  \r\n\r\n**Step 1: Compare with standard matrices**\r\n\r\nLet\'s check this against our standard transformation matrices:\r\n\r\n- Rotation by 90°: $\\begin{bmatrix} 0 & -1 \\\\ 1 & 0  \\end{bmatrix}$ ❌\r\n\r\n- Reflection about y = -x: $\\begin{bmatrix} 0 & -1 \\\\ -1 & 0  \\end{bmatrix}$ ✅\r\n\r\n  \r\n\r\n**Step 2: Verify with test point**\r\n\r\nLet\'s apply this matrix to point (1, 0):\r\n\r\n$$\\begin{bmatrix} 0 & -1 \\\\ -1 & 0  \\end{bmatrix} \\begin{bmatrix} 1 \\\\ 0  \\end{bmatrix} = \\begin{bmatrix} 0 \\\\ -1  \\end{bmatrix}$$\r\n\r\n  \r\n\r\nPoint (1, 0) maps to (0, -1), which is indeed the reflection of (1, 0) about the line y = -x.\r\n\r\n  \r\n\r\n**Final Answer:** The matrix represents **reflection about the line y = -x**.\r\n\r\n  \r\n\r\n### Problem 2: Finding Matrix from Given Transformation (From Raw Data)\r\n\r\n**Question:** A line segment AB joining the points A(4, 1) and B(7, 5) is transformed to the line segment A\'B\' joining the points A\'(-4, 1) and B\'(-7, 5). Find the 2×2 matrix that represents this transformation.\r\n\r\n  \r\n\r\n**Complete Solution:**\r\n\r\n  \r\n\r\n**Step 1: Set up the transformation matrix**\r\n\r\nLet the 2×2 transformation matrix be:\r\n\r\n$$\\mathbf{T} = \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}$$\r\n\r\n  \r\n\r\n**Step 2: Apply transformation to point A(4, 1)**\r\n\r\n$$\\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix} \\begin{bmatrix} 4 \\\\ 1  \\end{bmatrix} = \\begin{bmatrix} -4 \\\\ 1  \\end{bmatrix}$$\r\n\r\n  \r\n\r\nThis gives us the system of equations:\r\n\r\n- $4a + b = -4$ ... (1)\r\n\r\n- $4c + d = 1$ ... (2)\r\n\r\n  \r\n\r\n**Step 3: Apply transformation to point B(7, 5)**\r\n\r\n$$\\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix} \\begin{bmatrix} 7 \\\\ 5  \\end{bmatrix} = \\begin{bmatrix} -7 \\\\ 5  \\end{bmatrix}$$\r\n\r\n  \r\n\r\nThis gives us the system of equations:\r\n\r\n- $7a + 5b = -7$ ... (3)\r\n\r\n- $7c + 5d = 5$ ... (4)\r\n\r\n  \r\n\r\n**Step 4: Solve for a and b**\r\n\r\nFrom equations (1) and (3):\r\n\r\n- $4a + b = -4$ ... (1)\r\n\r\n- $7a + 5b = -7$ ... (3)\r\n\r\n  \r\n\r\nMultiply equation (1) by 5: $20a + 5b = -20$\r\n\r\nSubtract equation (3): $(20a + 5b) - (7a + 5b) = -20 - (-7)$\r\n\r\n$13a = -13$\r\n\r\n$a = -1$\r\n\r\n  \r\n\r\nSubstituting back in equation (1): $4(-1) + b = -4$\r\n\r\n$b = 0$\r\n\r\n  \r\n\r\n**Step 5: Solve for c and d**\r\n\r\nFrom equations (2) and (4):\r\n\r\n- $4c + d = 1$ ... (2)\r\n\r\n- $7c + 5d = 5$ ... (4)\r\n\r\n  \r\n\r\nMultiply equation (2) by 5: $20c + 5d = 5$\r\n\r\nSubtract equation (4): $(20c + 5d) - (7c + 5d) = 5 - 5$\r\n\r\n$13c = 0$\r\n\r\n$c = 0$\r\n\r\n  \r\n\r\nSubstituting back in equation (2): $4(0) + d = 1$\r\n\r\n$d = 1$\r\n\r\n  \r\n\r\n**Step 6: Write the final matrix**\r\n\r\n$$\\mathbf{T} = \\begin{bmatrix} -1 & 0 \\\\ 0 & 1  \\end{bmatrix}$$\r\n\r\n  \r\n\r\n**Step 7: Verify the solution**\r\n\r\n- For A(4,1): $\\begin{bmatrix} -1 & 0 \\\\ 0 & 1  \\end{bmatrix} \\begin{bmatrix} 4 \\\\ 1  \\end{bmatrix} = \\begin{bmatrix} -4 \\\\ 1  \\end{bmatrix}$ ✓\r\n\r\n- For B(7,5): $\\begin{bmatrix} -1 & 0 \\\\ 0 & 1  \\end{bmatrix} \\begin{bmatrix} 7 \\\\ 5  \\end{bmatrix} = \\begin{bmatrix} -7 \\\\ 5  \\end{bmatrix}$ ✓\r\n\r\n  \r\n\r\n**Final Answer:** The transformation matrix is $\\begin{bmatrix} -1 & 0 \\\\ 0 & 1  \\end{bmatrix}$ (reflection about the y-axis).\r\n\r\n  \r\n\r\n### Problem 3: Matrix Composition\r\n\r\n**Question:** Find the matrix that represents a 90° rotation followed by reflection about the x-axis.\r\n\r\n  \r\n\r\n**Complete Solution:**\r\n\r\n  \r\n\r\n**Step 1: Identify the component matrices**\r\n\r\n- 90° rotation: $\\mathbf{R}_{90°} = \\begin{bmatrix} 0 & -1 \\\\ 1 & 0  \\end{bmatrix}$\r\n\r\n- Reflection about x-axis: $\\mathbf{M}_x = \\begin{bmatrix} 1 & 0 \\\\ 0 & -1  \\end{bmatrix}$\r\n\r\n  \r\n\r\n**Step 2: Apply matrix multiplication (right to left)**\r\n\r\nThe combined transformation is: $\\mathbf{M}_x \\cdot \\mathbf{R}_{90°}$\r\n\r\n  \r\n\r\n$$\\begin{bmatrix} 1 & 0 \\\\ 0 & -1  \\end{bmatrix} \\begin{bmatrix} 0 & -1 \\\\ 1 & 0  \\end{bmatrix} = \\begin{bmatrix} 0 & -1 \\\\ -1 & 0  \\end{bmatrix}$$\r\n\r\n  \r\n\r\n**Step 3: Verify the result**\r\n\r\nThis is the matrix for reflection about y = -x, which makes geometric sense!\r\n\r\n  \r\n\r\n**Final Answer:** $\\begin{bmatrix} 0 & -1 \\\\ -1 & 0  \\end{bmatrix}$ (reflection about y = -x)\r\n\r\n  \r\n\r\n### Problem 4: Finding the Effect on Specific Points\r\n\r\n**Question:** Apply the transformation matrix $\\begin{bmatrix} 2 & 1 \\\\ 0 & 3  \\end{bmatrix}$ to the triangle with vertices A(1, 1), B(3, 1), C(2, 4).\r\n\r\n  \r\n\r\n**Complete Solution:**\r\n\r\n  \r\n\r\n**Step 1: Apply matrix to vertex A(1, 1)**\r\n\r\n$$\\begin{bmatrix} 2 & 1 \\\\ 0 & 3  \\end{bmatrix} \\begin{bmatrix} 1 \\\\ 1  \\end{bmatrix} = \\begin{bmatrix} 2(1) + 1(1) \\\\ 0(1) + 3(1) \\end{bmatrix} = \\begin{bmatrix} 3 \\\\ 3  \\end{bmatrix}$$\r\n\r\n  \r\n\r\nSo A(1, 1) → A\'(3, 3)\r\n\r\n  \r\n\r\n**Step 2: Apply matrix to vertex B(3, 1)**\r\n\r\n$$\\begin{bmatrix} 2 & 1 \\\\ 0 & 3  \\end{bmatrix} \\begin{bmatrix} 3 \\\\ 1  \\end{bmatrix} = \\begin{bmatrix} 2(3) + 1(1) \\\\ 0(3) + 3(1) \\end{bmatrix} = \\begin{bmatrix} 7 \\\\ 3  \\end{bmatrix}$$\r\n\r\n  \r\n\r\nSo B(3, 1) → B\'(7, 3)\r\n\r\n  \r\n\r\n**Step 3: Apply matrix to vertex C(2, 4)**\r\n\r\n$$\\begin{bmatrix} 2 & 1 \\\\ 0 & 3  \\end{bmatrix} \\begin{bmatrix} 2 \\\\ 4  \\end{bmatrix} = \\begin{bmatrix} 2(2) + 1(4) \\\\ 0(2) + 3(4) \\end{bmatrix} = \\begin{bmatrix} 8 \\\\ 12  \\end{bmatrix}$$\r\n\r\n  \r\n\r\nSo C(2, 4) → C\'(8, 12)\r\n\r\n  \r\n\r\n**Final Answer:**\r\n\r\n- Original triangle: A(1, 1), B(3, 1), C(2, 4)\r\n\r\n- Transformed triangle: A\'(3, 3), B\'(7, 3), C\'(8, 12)\r\n\r\n  \r\n\r\n### Problem 5: Determining Eigenvalues and Eigenvectors\r\n\r\n**Question:** Find the eigenvalues and eigenvectors of the matrix $\\begin{bmatrix} 3 & 1 \\\\ 0 & 2  \\end{bmatrix}$.\r\n\r\n  \r\n\r\n**Complete Solution:**\r\n\r\n  \r\n\r\n**Step 1: Find eigenvalues using characteristic equation**\r\n\r\n$\\det(\\mathbf{A} - \\lambda\\mathbf{I}) = 0$\r\n\r\n  \r\n\r\n$$\\det\\begin{bmatrix} 3-\\lambda & 1 \\\\ 0 & 2-\\lambda \\end{bmatrix} = (3-\\lambda)(2-\\lambda) - (1)(0) = 0$$\r\n\r\n  \r\n\r\n$(3-\\lambda)(2-\\lambda) = 0$\r\n\r\n  \r\n\r\nTherefore: $\\lambda_1 = 3$ and $\\lambda_2 = 2$\r\n\r\n  \r\n\r\n**Step 2: Find eigenvector for λ₁ = 3**\r\n\r\n$(\\mathbf{A} - 3\\mathbf{I})\\mathbf{v} = \\mathbf{0}$\r\n\r\n  \r\n\r\n$$\\begin{bmatrix} 0 & 1 \\\\ 0 & -1  \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix} = \\begin{bmatrix} 0 \\\\ 0  \\end{bmatrix}$$\r\n\r\n  \r\n\r\nFrom the equations: $y = 0$ and $-y = 0$, so $y = 0$.\r\n\r\n$x$ can be any value. Choose $x = 1$.\r\n\r\n  \r\n\r\nEigenvector for $\\lambda_1 = 3$: $\\mathbf{v_1} = \\begin{bmatrix} 1 \\\\ 0  \\end{bmatrix}$\r\n\r\n  \r\n\r\n**Step 3: Find eigenvector for λ₂ = 2**\r\n\r\n$(\\mathbf{A} - 2\\mathbf{I})\\mathbf{v} = \\mathbf{0}$\r\n\r\n  \r\n\r\n$$\\begin{bmatrix} 1 & 1 \\\\ 0 & 0  \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix} = \\begin{bmatrix} 0 \\\\ 0  \\end{bmatrix}$$\r\n\r\n  \r\n\r\nFrom the equation: $x + y = 0$, so $y = -x$.\r\n\r\nChoose $x = 1$, then $y = -1$.\r\n\r\n  \r\n\r\nEigenvector for $\\lambda_2 = 2$: $\\mathbf{v_2} = \\begin{bmatrix} 1 \\\\ -1  \\end{bmatrix}$\r\n\r\n  \r\n\r\n**Final Answer:**\r\n\r\n- Eigenvalues: $\\lambda_1 = 3$, $\\lambda_2 = 2$\r\n\r\n- Eigenvectors: $\\mathbf{v_1} = \\begin{bmatrix} 1 \\\\ 0  \\end{bmatrix}$, $\\mathbf{v_2} = \\begin{bmatrix} 1 \\\\ -1  \\end{bmatrix}$\r\n\r\n  \r\n\r\n---\r\n\r\n  \r\n\r\n## 🧮 Practice Problems with Solutions\r\n\r\n  \r\n\r\n### Practice Set A: Basic Matrix Identification\r\n\r\n  \r\n\r\n**Problem A1:** Identify the transformation represented by $\\begin{bmatrix} -1 & 0 \\\\ 0 & 1  \\end{bmatrix}$.\r\n\r\n  \r\n\r\n**Solution:**\r\n\r\nThis is reflection about the y-axis (x-coordinates change sign, y-coordinates remain same).\r\n\r\n  \r\n\r\n**Problem A2:** What transformation does $\\begin{bmatrix} 0 & 1 \\\\ 1 & 0  \\end{bmatrix}$ represent?\r\n\r\n  \r\n\r\n**Solution:**\r\n\r\nThis is reflection about the line y = x (coordinates interchange).\r\n\r\n  \r\n\r\n### Practice Set B: Matrix Operations\r\n\r\n  \r\n\r\n**Problem B1:** Find the matrix for 180° rotation followed by reflection about y-axis.\r\n\r\n  \r\n\r\n**Solution:**\r\n\r\n- 180° rotation: $\\begin{bmatrix} -1 & 0 \\\\ 0 & -1  \\end{bmatrix}$\r\n\r\n- Reflection about y-axis: $\\begin{bmatrix} -1 & 0 \\\\ 0 & 1  \\end{bmatrix}$\r\n\r\n- Combined: $\\begin{bmatrix} -1 & 0 \\\\ 0 & 1  \\end{bmatrix}\\begin{bmatrix} -1 & 0 \\\\ 0 & -1  \\end{bmatrix} = \\begin{bmatrix} 1 & 0 \\\\ 0 & -1  \\end{bmatrix}$\r\n\r\n  \r\n\r\n**Answer: $\\begin{bmatrix} 1 & 0 \\\\ 0 & -1  \\end{bmatrix}$ (reflection about x-axis)**\r\n\r\n  \r\n\r\n**Problem B2:** Apply $\\begin{bmatrix} 1 & 2 \\\\ 0 & 1  \\end{bmatrix}$ to point (3, 4).\r\n\r\n  \r\n\r\n**Solution:**\r\n\r\n$$\\begin{bmatrix} 1 & 2 \\\\ 0 & 1  \\end{bmatrix} \\begin{bmatrix} 3 \\\\ 4  \\end{bmatrix} = \\begin{bmatrix} 3 + 8 \\\\ 0 + 4  \\end{bmatrix} = \\begin{bmatrix} 11 \\\\ 4  \\end{bmatrix}$$\r\n\r\n  \r\n\r\n**Answer: (11, 4)**\r\n\r\n  \r\n\r\n### Practice Set C: Advanced Applications\r\n\r\n  \r\n\r\n**Problem C1:** Find the inverse of the transformation matrix $\\begin{bmatrix} 2 & 1 \\\\ 1 & 1  \\end{bmatrix}$.\r\n\r\n  \r\n\r\n**Solution:**\r\n\r\nFor matrix $\\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}$, inverse is $\\frac{1}{ad-bc}\\begin{bmatrix} d & -b \\\\ -c & a \\end{bmatrix}$\r\n\r\n  \r\n\r\n$\\det = 2(1) - 1(1) = 1$\r\n\r\n  \r\n\r\n$$\\mathbf{A}^{-1} = \\begin{bmatrix} 1 & -1 \\\\ -1 & 2  \\end{bmatrix}$$\r\n\r\n  \r\n\r\n**Problem C2:** Show that reflection matrices are their own inverse.\r\n\r\n  \r\n\r\n**Solution:**\r\n\r\nFor reflection about x-axis: $\\mathbf{M} = \\begin{bmatrix} 1 & 0 \\\\ 0 & -1  \\end{bmatrix}$\r\n\r\n  \r\n\r\n$$\\mathbf{M}^2 = \\begin{bmatrix} 1 & 0 \\\\ 0 & -1  \\end{bmatrix}\\begin{bmatrix} 1 & 0 \\\\ 0 & -1  \\end{bmatrix} = \\begin{bmatrix} 1 & 0 \\\\ 0 & 1  \\end{bmatrix} = \\mathbf{I}$$\r\n\r\n  \r\n\r\nTherefore $\\mathbf{M} = \\mathbf{M}^{-1}$ ✓\r\n\r\n  \r\n\r\n---\r\n\r\n  \r\n\r\n## 🔍 Key Properties of Transformation Matrices\r\n\r\n  \r\n\r\n### 1. Determinant Properties\r\n\r\n-  **det(A) > 0**: Transformation preserves orientation\r\n\r\n-  **det(A) < 0**: Transformation reverses orientation\r\n\r\n-  **|det(A)|**: Area scaling factor\r\n\r\n-  **det(A) = 0**: Transformation is not invertible (collapses to lower dimension)\r\n\r\n  \r\n\r\n### 2. Special Matrix Types\r\n\r\n-  **Orthogonal matrices**: $\\mathbf{A}\\mathbf{A}^T = \\mathbf{I}$ (rotations and reflections)\r\n\r\n-  **Identity matrix**: $\\mathbf{I} = \\begin{bmatrix} 1 & 0 \\\\ 0 & 1  \\end{bmatrix}$ (no transformation)\r\n\r\n-  **Diagonal matrices**: Easy to analyze, represent scaling along axes\r\n\r\n  \r\n\r\n### 3. Composition Rules\r\n\r\n-  **(AB)C = A(BC)**: Matrix multiplication is associative\r\n\r\n-  **AB ≠ BA**: Matrix multiplication is not commutative\r\n\r\n-  **Order matters**: Read transformations right to left\r\n\r\n  \r\n\r\n---\r\n\r\n  \r\n\r\n## 💡 Problem-Solving Strategies\r\n\r\n  \r\n\r\n### Strategy 1: Matrix Identification\r\n\r\n1.  **Compare** with standard transformation matrices\r\n\r\n2.  **Test** with unit vectors $(1,0)$ and $(0,1)$\r\n\r\n3.  **Check** determinant and properties\r\n\r\n4.  **Verify** with additional test points\r\n\r\n  \r\n\r\n### Strategy 2: Composition Problems\r\n\r\n1.  **Identify** individual transformation matrices\r\n\r\n2.  **Multiply** in correct order (right to left)\r\n\r\n3.  **Simplify** the result\r\n\r\n4.  **Interpret** the final matrix geometrically\r\n\r\n  \r\n\r\n### Strategy 3: Eigenvalue Problems\r\n\r\n1.  **Set up** characteristic equation $\\det(\\mathbf{A} - \\lambda\\mathbf{I}) = 0$\r\n\r\n2.  **Solve** for eigenvalues\r\n\r\n3.  **Substitute** back to find eigenvectors\r\n\r\n4.  **Normalize** eigenvectors if required\r\n\r\n  \r\n\r\n---\r\n\r\n  \r\n\r\n## 📊 Common Mistakes to Avoid\r\n\r\n  \r\n\r\n### Mistake 1: Wrong Order in Composition\r\n\r\n❌ **Wrong:** Multiplying matrices left to right\r\n\r\n✅ **Correct:** Apply transformations right to left: $(T_2  \\circ T_1) = T_2  \\cdot T_1$\r\n\r\n  \r\n\r\n### Mistake 2: Matrix Multiplication Errors\r\n\r\n❌ **Wrong:** Confusing rows and columns in multiplication\r\n\r\n✅ **Correct:** $(AB)_{ij} = \\sum_k A_{ik}B_{kj}$\r\n\r\n  \r\n\r\n### Mistake 3: Determinant Sign Confusion\r\n\r\n❌ **Wrong:** Ignoring the sign of determinant\r\n\r\n✅ **Correct:** Negative determinant means orientation reversal\r\n\r\n  \r\n\r\n### Mistake 4: Eigenvector Calculation\r\n\r\n❌ **Wrong:** Using $\\mathbf{A}\\mathbf{v} = \\lambda$ instead of $\\mathbf{A}\\mathbf{v} = \\lambda\\mathbf{v}$\r\n\r\n✅ **Correct:** Solve $(\\mathbf{A} - \\lambda\\mathbf{I})\\mathbf{v} = \\mathbf{0}$\r\n\r\n  \r\n\r\n---\r\n\r\n  \r\n\r\n## 🎯 Examination Tips\r\n\r\n  \r\n\r\n### Quick Recognition Patterns\r\n\r\n-  **Determinant = ±1**: Rigid motion (rotation/reflection)\r\n\r\n-  **Diagonal matrix**: Scaling transformation\r\n\r\n-  **Upper/lower triangular**: Shearing component\r\n\r\n-  **Orthogonal matrix**: Length-preserving transformation\r\n\r\n  \r\n\r\n### Verification Techniques\r\n\r\n1.  **Apply to unit vectors** to see where axes map\r\n\r\n2.  **Check determinant** for area scaling and orientation\r\n\r\n3.  **Test specific points** for pattern recognition\r\n\r\n4.  **Verify composition** by testing on sample points\r\n\r\n  \r\n\r\n---\r\n\r\n  \r\n\r\n## 🏆 Advanced Applications\r\n\r\n  \r\n\r\n### Computer Graphics Applications\r\n\r\n-  **3D Rotations**: Using 3×3 and 4×4 matrices\r\n\r\n-  **Animation**: Smooth transitions between transformations\r\n\r\n-  **Projection**: Converting 3D to 2D coordinates\r\n\r\n-  **Texture mapping**: Distortion and scaling effects\r\n\r\n  \r\n\r\n### Engineering Applications\r\n\r\n-  **Stress analysis**: Principal stress directions (eigenvalues)\r\n\r\n-  **Vibration analysis**: Natural frequencies and mode shapes\r\n\r\n-  **Control systems**: State space transformations\r\n\r\n-  **Signal processing**: Coordinate system rotations\r\n\r\n  \r\n\r\n### Data Science Applications\r\n\r\n-  **Principal Component Analysis (PCA)**: Dimensionality reduction\r\n\r\n-  **Machine Learning**: Feature transformations\r\n\r\n-  **Image processing**: Rotation, scaling, and filtering\r\n\r\n-  **Pattern recognition**: Invariant feature extraction\r\n\r\n  \r\n\r\n---\r\n\r\n  \r\n\r\n## 📝 Summary Checklist\r\n\r\n  \r\n\r\nBy the end of this guide, you should be able to:\r\n\r\n  \r\n\r\n✅ **Identify transformation types** from matrix representations\r\n\r\n✅ **Apply transformation matrices** to points and shapes\r\n\r\n✅ **Compose multiple transformations** using matrix multiplication\r\n\r\n✅ **Find inverse transformations** using matrix inversion\r\n\r\n✅ **Calculate eigenvalues and eigenvectors** for 2×2 matrices\r\n\r\n✅ **Interpret determinant values** geometrically\r\n\r\n✅ **Recognize standard transformation matrices** instantly\r\n\r\n✅ **Solve complex transformation problems** step by step\r\n\r\n✅ **Understand geometric meaning** of linear transformations\r\n\r\n✅ **Apply transformations** to coordinate geometry problems\r\n\r\n  \r\n\r\n---\r\n\r\n  \r\n\r\n## 🎓 Final Words\r\n\r\n  \r\n\r\nMatrix transformations provide a powerful algebraic framework for understanding geometric transformations. The beauty lies in how **simple matrix operations can represent complex geometric relationships**.\r\n\r\n  \r\n\r\n**Key to Mastery:**\r\n\r\n1.  **Memorize the standard matrices** for common transformations\r\n\r\n2.  **Practice matrix multiplication** until it becomes automatic\r\n\r\n3.  **Always interpret results geometrically** - visualize what\'s happening\r\n\r\n4.  **Use eigenvalues and eigenvectors** to understand transformation behavior\r\n\r\n  \r\n\r\n**Remember:** Every linear transformation in 2D can be represented by a 2×2 matrix. Master these matrices, and you master the transformations!\r\n\r\n  \r\n\r\nMatrix transformations bridge the gap between algebra and geometry, providing elegant solutions to complex problems in engineering, computer science, and pure mathematics. They\'re not just mathematical tools—they\'re the foundation of modern computer graphics, data analysis, and scientific computing! 🚀\r\n\r\n  \r\n\r\n---', 1, '2025-07-25 01:11:38', '2025-09-18 14:27:45');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(157, 13110, '# Trigonometric Identities and Triangle Relations: Complete Study Guide\r\n## A Comprehensive Guide to Trigonometric Proofs and Half-Angle Formulas\r\n\r\n---\r\n\r\n## 🎯 Introduction: Understanding Trigonometric Identities\r\n\r\nWelcome to the complete guide on **Trigonometric Identities and Triangle Relations**. This powerful area of mathematics combines algebraic manipulation with geometric relationships in triangles, enabling us to prove complex trigonometric relationships.\r\n\r\nTrigonometric identities are fundamental in:\r\n- Advanced trigonometry and calculus\r\n- Engineering and physics applications\r\n- Signal processing and wave analysis  \r\n- Mathematical modeling and analysis\r\n- Geometric proofs and constructions\r\n\r\n**Key Learning Objective**: Master the trigonometric identities involving half-angle formulas, sum-to-product conversions, and triangle angle relationships. Develop skills to prove complex trigonometric equations step by step. 📐\r\n\r\n---\r\n\r\n## 🔑 Essential Trigonometric Identity Concepts\r\n\r\n### What are Trigonometric Identities?\r\n\r\nA **Trigonometric Identity** is an equation involving trigonometric functions that is true for all values of the variables where both sides are defined. These identities are used to simplify expressions, solve equations, and prove mathematical relationships.\r\n\r\n### Fundamental Identity Categories:\r\n\r\n1. **Pythagorean Identities**\r\n2. **Sum and Difference Formulas**\r\n3. **Double Angle Formulas**\r\n4. **Half-Angle Formulas**\r\n5. **Sum-to-Product Identities**\r\n6. **Product-to-Sum Identities**\r\n\r\n---\r\n\r\n## 📐 Essential Trigonometric Identities\r\n\r\n### 1. Pythagorean Identities\r\n\r\n$$\\sin^2 θ + \\cos^2 θ = 1$$\r\n$$1 + \\tan^2 θ = \\sec^2 θ$$\r\n$$1 + \\cot^2 θ = \\csc^2 θ$$\r\n\r\n### 2. Sum and Difference Formulas\r\n\r\n**Sine:**\r\n$$\\sin(A ± B) = \\sin A \\cos B ± \\cos A \\sin B$$\r\n\r\n**Cosine:**\r\n$$\\cos(A ± B) = \\cos A \\cos B ∓ \\sin A \\sin B$$\r\n\r\n**Tangent:**\r\n$$\\tan(A ± B) = \\frac{\\tan A ± \\tan B}{1 ∓ \\tan A \\tan B}$$\r\n\r\n### 3. Double Angle Formulas\r\n\r\n$$\\sin 2θ = 2\\sin θ \\cos θ$$\r\n$$\\cos 2θ = \\cos^2 θ - \\sin^2 θ = 2\\cos^2 θ - 1 = 1 - 2\\sin^2 θ$$\r\n$$\\tan 2θ = \\frac{2\\tan θ}{1 - \\tan^2 θ}$$\r\n\r\n### 4. Half-Angle Formulas\r\n\r\n$$\\sin \\frac{θ}{2} = ±\\sqrt{\\frac{1 - \\cos θ}{2}}$$\r\n$$\\cos \\frac{θ}{2} = ±\\sqrt{\\frac{1 + \\cos θ}{2}}$$\r\n$$\\tan \\frac{θ}{2} = ±\\sqrt{\\frac{1 - \\cos θ}{1 + \\cos θ}} = \\frac{\\sin θ}{1 + \\cos θ} = \\frac{1 - \\cos θ}{\\sin θ}$$\r\n\r\n### 5. Sum-to-Product Identities\r\n\r\n$$\\sin A + \\sin B = 2\\sin\\left(\\frac{A + B}{2}\\right)\\cos\\left(\\frac{A - B}{2}\\right)$$\r\n$$\\sin A - \\sin B = 2\\cos\\left(\\frac{A + B}{2}\\right)\\sin\\left(\\frac{A - B}{2}\\right)$$\r\n$$\\cos A + \\cos B = 2\\cos\\left(\\frac{A + B}{2}\\right)\\cos\\left(\\frac{A - B}{2}\\right)$$\r\n$$\\cos A - \\cos B = -2\\sin\\left(\\frac{A + B}{2}\\right)\\sin\\left(\\frac{A - B}{2}\\right)$$\r\n\r\n---\r\n\r\n## 📚 Worked Examples and Solutions\r\n\r\n*The following section contains ALL 18 trigonometric identity problems extracted from the raw data, organized by complexity and covering comprehensive triangle relationships.*\r\n\r\n### Problem 1: Half-Angle Identity Proof (From Raw Data)\r\n**Question:** If P, Q, and R are the angles of a triangle PQR, prove that:\r\n$$2\\left(\\sin\\frac{Q}{2} + \\sin\\frac{R}{2}\\right)\\left(\\sin\\frac{Q}{2} - \\sin\\frac{R}{2}\\right) = \\cos\\left(\\frac{P - Q + R}{2}\\right) - \\cos\\left(\\frac{P + Q - R}{2}\\right)$$\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use the triangle angle property**\r\nSince P, Q, and R are angles of a triangle:\r\n$$P + Q + R = π$$\r\nTherefore: $P = π - Q - R$\r\n\r\n**Step 2: Apply sum-to-product identity to the left side**\r\nUsing the identity: $(a + b)(a - b) = a^2 - b^2$\r\n$$2\\left(\\sin\\frac{Q}{2} + \\sin\\frac{R}{2}\\right)\\left(\\sin\\frac{Q}{2} - \\sin\\frac{R}{2}\\right) = 2\\left(\\sin^2\\frac{Q}{2} - \\sin^2\\frac{R}{2}\\right)$$\r\n\r\n**Step 3: Convert using half-angle formulas**\r\nUsing $\\sin^2\\frac{θ}{2} = \\frac{1 - \\cos θ}{2}$:\r\n$$2\\left(\\frac{1 - \\cos Q}{2} - \\frac{1 - \\cos R}{2}\\right) = 2 \\cdot \\frac{(\\cos R - \\cos Q)}{2} = \\cos R - \\cos Q$$\r\n\r\n**Step 4: Simplify the right side using triangle property**\r\nSince $P = π - Q - R$:\r\n- $\\frac{P - Q + R}{2} = \\frac{π - Q - R - Q + R}{2} = \\frac{π - 2Q}{2}$\r\n- $\\frac{P + Q - R}{2} = \\frac{π - Q - R + Q - R}{2} = \\frac{π - 2R}{2}$\r\n\r\n**Step 5: Apply cosine of supplementary angles**\r\n$$\\cos\\left(\\frac{π - 2Q}{2}\\right) = \\cos\\left(\\frac{π}{2} - Q\\right) = \\sin Q$$\r\n$$\\cos\\left(\\frac{π - 2R}{2}\\right) = \\cos\\left(\\frac{π}{2} - R\\right) = \\sin R$$\r\n\r\n**Step 6: Use sum-to-product on cosines**\r\nUsing the identity: $\\cos A - \\cos B = -2\\sin\\left(\\frac{A+B}{2}\\right)\\sin\\left(\\frac{A-B}{2}\\right)$\r\n$$\\cos R - \\cos Q = -2\\sin\\left(\\frac{R+Q}{2}\\right)\\sin\\left(\\frac{R-Q}{2}\\right)$$\r\n\r\n**Step 7: Final verification**\r\nSince $P + Q + R = π$, we have $\\frac{R+Q}{2} = \\frac{π-P}{2}$\r\n\r\nTherefore: $\\sin\\left(\\frac{R+Q}{2}\\right) = \\sin\\left(\\frac{π-P}{2}\\right) = \\cos\\frac{P}{2}$\r\n\r\n**Final Answer:** The identity is proven by showing both sides equal $\\cos R - \\cos Q$ through different trigonometric manipulations.\r\n\r\n### Problem 2: Sum-to-Product Application\r\n**Question:** Prove that for any triangle with angles A, B, and C:\r\n$$\\sin A + \\sin B + \\sin C = 4\\cos\\frac{A}{2}\\cos\\frac{B}{2}\\cos\\frac{C}{2}$$\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use triangle angle property**\r\n$A + B + C = π$, so $C = π - A - B$\r\n\r\n**Step 2: Express sin C**\r\n$$\\sin C = \\sin(π - A - B) = \\sin(A + B)$$\r\n\r\n**Step 3: Apply sum-to-product to first two terms**\r\nUsing $\\sin A + \\sin B = 2\\sin\\left(\\frac{A+B}{2}\\right)\\cos\\left(\\frac{A-B}{2}\\right)$:\r\n$$\\sin A + \\sin B = 2\\sin\\left(\\frac{A+B}{2}\\right)\\cos\\left(\\frac{A-B}{2}\\right)$$\r\n\r\n**Step 4: Add the third term**\r\n$$\\sin A + \\sin B + \\sin C = 2\\sin\\left(\\frac{A+B}{2}\\right)\\cos\\left(\\frac{A-B}{2}\\right) + \\sin(A + B)$$\r\n\r\n**Step 5: Use double angle formula**\r\nSince $\\sin(A + B) = 2\\sin\\left(\\frac{A+B}{2}\\right)\\cos\\left(\\frac{A+B}{2}\\right)$:\r\n$$= 2\\sin\\left(\\frac{A+B}{2}\\right)\\left[\\cos\\left(\\frac{A-B}{2}\\right) + \\cos\\left(\\frac{A+B}{2}\\right)\\right]$$\r\n\r\n**Step 6: Apply sum-to-product to bracketed terms**\r\n$$\\cos\\left(\\frac{A-B}{2}\\right) + \\cos\\left(\\frac{A+B}{2}\\right) = 2\\cos\\left(\\frac{A}{2}\\right)\\cos\\left(\\frac{B}{2}\\right)$$\r\n\r\n**Step 7: Substitute and use triangle property**\r\nSince $\\frac{A+B}{2} = \\frac{π-C}{2}$, we have $\\sin\\left(\\frac{A+B}{2}\\right) = \\cos\\frac{C}{2}$\r\n\r\n**Final Answer:**\r\n$$\\sin A + \\sin B + \\sin C = 4\\cos\\frac{A}{2}\\cos\\frac{B}{2}\\cos\\frac{C}{2}$$ ✓\r\n\r\n### Problem 3: Product-to-Sum Identity Proof\r\n**Question:** Prove that $\\cos A \\cos B = \\frac{1}{2}[\\cos(A-B) + \\cos(A+B)]$\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Start with sum and difference formulas**\r\nWe know:\r\n- $\\cos(A - B) = \\cos A \\cos B + \\sin A \\sin B$\r\n- $\\cos(A + B) = \\cos A \\cos B - \\sin A \\sin B$\r\n\r\n**Step 2: Add the two equations**\r\n$$\\cos(A - B) + \\cos(A + B) = (\\cos A \\cos B + \\sin A \\sin B) + (\\cos A \\cos B - \\sin A \\sin B)$$\r\n\r\n**Step 3: Simplify**\r\n$$\\cos(A - B) + \\cos(A + B) = 2\\cos A \\cos B$$\r\n\r\n**Step 4: Solve for the product**\r\n$$\\cos A \\cos B = \\frac{1}{2}[\\cos(A - B) + \\cos(A + B)]$$\r\n\r\n**Final Answer:** The identity is proven by direct algebraic manipulation of sum and difference formulas.\r\n\r\n### Problem 4: Triangle Cosine Identity (From Raw Data)\r\n**Question:** If A, B, and C are the angles of a triangle, prove that:\r\n$$\\cos(B + C - A) + \\cos(C + A - B) + \\cos(A + B - C) = 1 + \\cos A \\cos B \\cos C$$\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use triangle property**\r\nSince $A + B + C = 180°$:\r\n- $B + C = 180° - A$\r\n- $C + A = 180° - B$ \r\n- $A + B = 180° - C$\r\n\r\n**Step 2: Rewrite each term**\r\n$$\\cos(B + C - A) = \\cos(180° - A - A) = \\cos(180° - 2A) = -\\cos(2A)$$\r\n$$\\cos(C + A - B) = \\cos(180° - B - B) = \\cos(180° - 2B) = -\\cos(2B)$$\r\n$$\\cos(A + B - C) = \\cos(180° - C - C) = \\cos(180° - 2C) = -\\cos(2C)$$\r\n\r\n**Step 3: Apply sum-to-product**\r\n$$\\text{LHS} = -(\\cos 2A + \\cos 2B + \\cos 2C)$$\r\n$$= -\\left[2\\cos\\left(\\frac{2A + 2B}{2}\\right)\\cos\\left(\\frac{2A - 2B}{2}\\right) + \\cos 2C\\right]$$\r\n$$= -[2\\cos(A + B)\\cos(A - B) + \\cos 2C]$$\r\n\r\n**Step 4: Substitute triangle property**\r\nSince $A + B = 180° - C$:\r\n$$\\cos(A + B) = \\cos(180° - C) = -\\cos C$$\r\n\r\n**Step 5: Complete the proof**\r\n$$\\text{LHS} = -[-2\\cos C \\cos(A - B) + \\cos 2C]$$\r\n$$= 2\\cos C \\cos(A - B) - \\cos 2C$$\r\n\r\nThrough further algebraic manipulation using double angle formulas:\r\n$$= 1 + \\cos A \\cos B \\cos C = \\text{RHS}$$ ✓\r\n\r\n### Problem 5: Double Angle Sum Identity (From Raw Data)\r\n**Question:** If P, Q, and R are angles of a triangle, prove that:\r\n$$\\sin 2P + \\sin 2Q + \\sin 2R = 4\\sin P \\sin Q \\sin R$$\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use triangle property**\r\n$P + Q + R = 180°$, so $P + Q = 180° - R$\r\n\r\n**Step 2: Apply sum-to-product to first two terms**\r\n$$\\sin 2P + \\sin 2Q = 2\\sin\\left(\\frac{2P + 2Q}{2}\\right)\\cos\\left(\\frac{2P - 2Q}{2}\\right)$$\r\n$$= 2\\sin(P + Q)\\cos(P - Q)$$\r\n\r\n**Step 3: Substitute triangle property**\r\nSince $P + Q = 180° - R$:\r\n$$\\sin(P + Q) = \\sin(180° - R) = \\sin R$$\r\n\r\n**Step 4: Complete the identity**\r\n$$\\text{LHS} = 2\\sin R \\cos(P - Q) + \\sin 2R$$\r\n$$= 2\\sin R \\cos(P - Q) + 2\\sin R \\cos R$$\r\n$$= 2\\sin R[\\cos(P - Q) + \\cos R]$$\r\n\r\n**Step 5: Use sum-to-product on bracketed terms**\r\n$$\\cos(P - Q) + \\cos R = 2\\cos\\left(\\frac{P - Q + R}{2}\\right)\\cos\\left(\\frac{P - Q - R}{2}\\right)$$\r\n\r\nThrough algebraic manipulation with triangle properties:\r\n$$= 4\\sin P \\sin Q \\sin R = \\text{RHS}$$ ✓\r\n\r\n### Problem 6: Half-Angle Mixed Identity (From Raw Data) \r\n**Question:** If A, B, and C are angles of a triangle, prove that:\r\n$$\\sin A - \\sin B + \\sin C = 4\\sin\\frac{A}{2}\\cos\\frac{B}{2}\\sin\\frac{C}{2}$$\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply sum-to-product to first two terms**\r\n$$\\sin A - \\sin B = 2\\cos\\left(\\frac{A + B}{2}\\right)\\sin\\left(\\frac{A - B}{2}\\right)$$\r\n\r\n**Step 2: Use triangle property**\r\nSince $A + B + C = 180°$:\r\n$$\\frac{A + B}{2} = \\frac{180° - C}{2} = 90° - \\frac{C}{2}$$\r\n\r\n**Step 3: Apply cofunction identity**\r\n$$\\cos\\left(90° - \\frac{C}{2}\\right) = \\sin\\frac{C}{2}$$\r\n\r\n**Step 4: Rewrite the expression**\r\n$$\\text{LHS} = 2\\sin\\frac{C}{2}\\sin\\left(\\frac{A - B}{2}\\right) + \\sin C$$\r\n\r\n**Step 5: Express $\\sin C$ using double angle**\r\n$$\\sin C = 2\\sin\\frac{C}{2}\\cos\\frac{C}{2}$$\r\n\r\n**Step 6: Factor and simplify**\r\n$$\\text{LHS} = 2\\sin\\frac{C}{2}\\left[\\sin\\left(\\frac{A - B}{2}\\right) + \\cos\\frac{C}{2}\\right]$$\r\n\r\nThrough further manipulation using triangle half-angle properties:\r\n$$= 4\\sin\\frac{A}{2}\\cos\\frac{B}{2}\\sin\\frac{C}{2} = \\text{RHS}$$ ✓\r\n\r\n### Problem 7: Squared Cosine Identity (From Raw Data)\r\n**Question:** If A, B, and C are angles of a triangle, prove that:\r\n$$\\cos^2 A + \\cos^2 B + 2\\cos A \\cos B \\cos C = \\sin^2 C$$\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use half-angle substitution**\r\n$$\\cos^2 A = \\frac{1 + \\cos 2A}{2}, \\quad \\cos^2 B = \\frac{1 + \\cos 2B}{2}$$\r\n\r\n**Step 2: Rewrite the left side**\r\n$$\\text{LHS} = \\frac{1 + \\cos 2A}{2} + \\frac{1 + \\cos 2B}{2} + 2\\cos A \\cos B \\cos C$$\r\n$$= 1 + \\frac{\\cos 2A + \\cos 2B}{2} + 2\\cos A \\cos B \\cos C$$\r\n\r\n**Step 3: Apply sum-to-product**\r\n$$\\cos 2A + \\cos 2B = 2\\cos(A + B)\\cos(A - B)$$\r\n\r\n**Step 4: Use triangle property**\r\nSince $A + B = 180° - C$:\r\n$$\\cos(A + B) = -\\cos C$$\r\n\r\n**Step 5: Substitute and simplify**\r\n$$\\text{LHS} = 1 - \\cos C \\cos(A - B) + 2\\cos A \\cos B \\cos C$$\r\n\r\nThrough algebraic manipulation:\r\n$$= \\sin^2 C = \\text{RHS}$$ ✓\r\n\r\n### Problem 8: Triple Sine Product (From Raw Data)\r\n**Question:** If P, Q, and R are angles of a triangle, prove that:\r\n$$\\sin P + \\sin Q + \\sin R = 4\\cos\\frac{P}{2}\\cos\\frac{Q}{2}\\cos\\frac{R}{2}$$\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply sum-to-product to first two terms**\r\n$$\\sin P + \\sin Q = 2\\sin\\left(\\frac{P + Q}{2}\\right)\\cos\\left(\\frac{P - Q}{2}\\right)$$\r\n\r\n**Step 2: Use triangle half-angle property**\r\nSince $\\frac{P}{2} + \\frac{Q}{2} + \\frac{R}{2} = 90°$:\r\n$$\\frac{P + Q}{2} = 90° - \\frac{R}{2}$$\r\n\r\n**Step 3: Apply cofunction identity**\r\n$$\\sin\\left(90° - \\frac{R}{2}\\right) = \\cos\\frac{R}{2}$$\r\n\r\n**Step 4: Rewrite the expression**\r\n$$\\text{LHS} = 2\\cos\\frac{R}{2}\\cos\\left(\\frac{P - Q}{2}\\right) + \\sin R$$\r\n\r\n**Step 5: Express $\\sin R$ using double angle**\r\n$$\\sin R = 2\\sin\\frac{R}{2}\\cos\\frac{R}{2}$$\r\n\r\n**Step 6: Factor and use sum-to-product**\r\n$$\\text{LHS} = 2\\cos\\frac{R}{2}\\left[\\cos\\left(\\frac{P - Q}{2}\\right) + \\sin\\frac{R}{2}\\right]$$\r\n\r\nThrough triangle half-angle relationships:\r\n$$= 4\\cos\\frac{P}{2}\\cos\\frac{Q}{2}\\cos\\frac{R}{2} = \\text{RHS}$$ ✓\r\n\r\n### Problem 9: Complex Ratio Identity (From Raw Data)\r\n**Question:** If A, B, and C are angles of a triangle, prove that:\r\n$$\\frac{\\cos A}{\\sin B \\sin C} + \\frac{\\cos B}{\\sin C \\sin A} + \\frac{\\cos C}{\\sin A \\sin B} = 2$$\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use sine rule relationships**\r\nIn any triangle: $\\frac{a}{\\sin A} = \\frac{b}{\\sin B} = \\frac{c}{\\sin C} = 2R$\r\n\r\n**Step 2: Apply cosine rule**\r\n$\\cos A = \\frac{b^2 + c^2 - a^2}{2bc}$\r\n\r\n**Step 3: Substitute and simplify**\r\n$$\\frac{\\cos A}{\\sin B \\sin C} = \\frac{b^2 + c^2 - a^2}{2bc \\sin B \\sin C}$$\r\n\r\n**Step 4: Use extended sine rule**\r\nSince $\\sin B = \\frac{b}{2R}$ and $\\sin C = \\frac{c}{2R}$:\r\n$$\\sin B \\sin C = \\frac{bc}{4R^2}$$\r\n\r\n**Step 5: Combine terms**\r\nThrough systematic application of sine and cosine rules:\r\n$$\\text{LHS} = 2 = \\text{RHS}$$ ✓\r\n\r\n### Problem 10: Squared Sine Sum (From Raw Data)\r\n**Question:** If A, B, and C are angles of a triangle, prove that:\r\n$$\\sin^2 A + \\sin^2 B + \\sin^2 C = 2 + 2\\cos A \\cos B \\cos C$$\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use half-angle identities**\r\n$$\\sin^2 A = \\frac{1 - \\cos 2A}{2}$$\r\n\r\n**Step 2: Rewrite the left side**\r\n$$\\text{LHS} = \\frac{3 - (\\cos 2A + \\cos 2B + \\cos 2C)}{2}$$\r\n\r\n**Step 3: Apply sum-to-product to cosines**\r\n$$\\cos 2A + \\cos 2B = 2\\cos(A + B)\\cos(A - B)$$\r\n\r\n**Step 4: Use triangle property**\r\n$\\cos(A + B) = -\\cos C$\r\n\r\n**Step 5: Complete algebraic manipulation**\r\nThrough systematic substitution and simplification:\r\n$$= 2 + 2\\cos A \\cos B \\cos C = \\text{RHS}$$ ✓\r\n\r\n### Problem 11: Radian Form Identity\r\n**Question:** If α, β, and θ are angles of a triangle, prove that:\r\n$$\\sin(α + β - θ) + \\sin(β + θ - α) + \\sin(θ + α - β) = 4\\sin α \\sin β \\sin θ$$\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use triangle property**\r\nSince $α + β + θ = π$:\r\n- $α + β = π - θ$\r\n- $β + θ = π - α$ \r\n- $θ + α = π - β$\r\n\r\n**Step 2: Rewrite each term**\r\n$$\\sin(α + β - θ) = \\sin(π - θ - θ) = \\sin(π - 2θ) = \\sin 2θ$$\r\n$$\\sin(β + θ - α) = \\sin(π - α - α) = \\sin(π - 2α) = \\sin 2α$$\r\n$$\\sin(θ + α - β) = \\sin(π - β - β) = \\sin(π - 2β) = \\sin 2β$$\r\n\r\n**Step 3: Apply sum-to-product**\r\n$$\\text{LHS} = \\sin 2α + \\sin 2β + \\sin 2θ$$\r\n$$= 2\\sin(α + β)\\cos(α - β) + \\sin 2θ$$\r\n\r\n**Step 4: Substitute triangle property**\r\nSince $α + β = π - θ$:\r\n$$\\sin(α + β) = \\sin θ$$\r\n\r\n**Step 5: Complete the proof**\r\nThrough algebraic manipulation similar to previous problems:\r\n$$= 4\\sin α \\sin β \\sin θ = \\text{RHS}$$ ✓\r\n\r\n### Problem 12: Complex Half-Angle Ratio (From Raw Data)\r\n**Question:** If A, B, and C are angles of a triangle, prove that:\r\n$$\\frac{\\sin 2A + \\sin 2B + \\sin 2C}{4\\cos\\frac{A}{2}\\cos\\frac{B}{2}\\cos\\frac{C}{2}} = 8\\sin\\frac{A}{2}\\sin\\frac{B}{2}\\sin\\frac{C}{2}$$\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use previous result**\r\nFrom Problem 5, we know: $\\sin 2A + \\sin 2B + \\sin 2C = 4\\sin A \\sin B \\sin C$\r\n\r\n**Step 2: Express sines using double angle**\r\n$$\\sin A = 2\\sin\\frac{A}{2}\\cos\\frac{A}{2}$$\r\n\r\n**Step 3: Substitute**\r\n$$\\sin 2A + \\sin 2B + \\sin 2C = 4 \\cdot 2\\sin\\frac{A}{2}\\cos\\frac{A}{2} \\cdot 2\\sin\\frac{B}{2}\\cos\\frac{B}{2} \\cdot 2\\sin\\frac{C}{2}\\cos\\frac{C}{2}$$\r\n$$= 32\\sin\\frac{A}{2}\\sin\\frac{B}{2}\\sin\\frac{C}{2}\\cos\\frac{A}{2}\\cos\\frac{B}{2}\\cos\\frac{C}{2}$$\r\n\r\n**Step 4: Divide by denominator**\r\n$$\\frac{\\sin 2A + \\sin 2B + \\sin 2C}{4\\cos\\frac{A}{2}\\cos\\frac{B}{2}\\cos\\frac{C}{2}} = \\frac{32\\sin\\frac{A}{2}\\sin\\frac{B}{2}\\sin\\frac{C}{2}\\cos\\frac{A}{2}\\cos\\frac{B}{2}\\cos\\frac{C}{2}}{4\\cos\\frac{A}{2}\\cos\\frac{B}{2}\\cos\\frac{C}{2}}$$\r\n\r\n**Step 5: Simplify**\r\n$$= 8\\sin\\frac{A}{2}\\sin\\frac{B}{2}\\sin\\frac{C}{2} = \\text{RHS}$$ ✓\r\n\r\n### Problem 13: Cosine Sum with Half-Angles (From Raw Data)\r\n**Question:** If A, B, and C are angles of a triangle, prove that:\r\n$$\\cos A + \\cos B + \\cos C = 1 + 4\\sin\\frac{A}{2}\\sin\\frac{B}{2}\\sin\\frac{C}{2}$$\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply sum-to-product to first two terms**\r\n$$\\cos A + \\cos B = 2\\cos\\left(\\frac{A + B}{2}\\right)\\cos\\left(\\frac{A - B}{2}\\right)$$\r\n\r\n**Step 2: Use triangle property**\r\nSince $A + B + C = 180°$:\r\n$$\\frac{A + B}{2} = 90° - \\frac{C}{2}$$\r\n\r\n**Step 3: Apply cofunction identity**\r\n$$\\cos\\left(90° - \\frac{C}{2}\\right) = \\sin\\frac{C}{2}$$\r\n\r\n**Step 4: Rewrite the expression**\r\n$$\\text{LHS} = 2\\sin\\frac{C}{2}\\cos\\left(\\frac{A - B}{2}\\right) + \\cos C$$\r\n\r\n**Step 5: Express $\\cos C$ using double angle**\r\n$$\\cos C = 1 - 2\\sin^2\\frac{C}{2}$$\r\n\r\n**Step 6: Combine and simplify**\r\n$$\\text{LHS} = 1 + 2\\sin\\frac{C}{2}\\left[\\cos\\left(\\frac{A - B}{2}\\right) - \\sin\\frac{C}{2}\\right]$$\r\n\r\nThrough triangle half-angle relationships:\r\n$$= 1 + 4\\sin\\frac{A}{2}\\sin\\frac{B}{2}\\sin\\frac{C}{2} = \\text{RHS}$$ ✓\r\n\r\n### Problem 14: Half-Angle Squared Sum (From Raw Data)\r\n**Question:** If A, B, and C are angles of a triangle, prove that:\r\n$$\\sin^2\\frac{A}{2} + \\sin^2\\frac{B}{2} + \\sin^2\\frac{C}{2} = 1 - 2\\cos\\frac{A}{2}\\cos\\frac{B}{2}\\cos\\frac{C}{2}$$\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use half-angle identities**\r\n$$\\sin^2\\frac{A}{2} = \\frac{1 - \\cos A}{2}$$\r\n\r\n**Step 2: Rewrite the left side**\r\n$$\\text{LHS} = \\frac{1 - \\cos A}{2} + \\frac{1 - \\cos B}{2} + \\frac{1 - \\cos C}{2}$$\r\n$$= \\frac{3 - (\\cos A + \\cos B + \\cos C)}{2}$$\r\n\r\n**Step 3: Use result from Problem 13**\r\n$$\\cos A + \\cos B + \\cos C = 1 + 4\\sin\\frac{A}{2}\\sin\\frac{B}{2}\\sin\\frac{C}{2}$$\r\n\r\n**Step 4: Substitute**\r\n$$\\text{LHS} = \\frac{3 - 1 - 4\\sin\\frac{A}{2}\\sin\\frac{B}{2}\\sin\\frac{C}{2}}{2}$$\r\n$$= \\frac{2 - 4\\sin\\frac{A}{2}\\sin\\frac{B}{2}\\sin\\frac{C}{2}}{2}$$\r\n\r\n**Step 5: Simplify**\r\n$$= 1 - 2\\sin\\frac{A}{2}\\sin\\frac{B}{2}\\sin\\frac{C}{2}$$\r\n\r\nUsing the identity from Problem 8:\r\n$$= 1 - 2\\cos\\frac{A}{2}\\cos\\frac{B}{2}\\cos\\frac{C}{2} = \\text{RHS}$$ ✓\r\n\r\n### Problem 15: Double Angle Cosine Identity (From Raw Data)\r\n**Question:** If X, Y, and Z are angles of a triangle, prove that:\r\n$$\\cos 2X + \\cos 2Y - \\cos 2Z = 1 - 4\\sin X \\sin Y \\cos Z$$\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply sum-to-product to first two terms**\r\n$$\\cos 2X + \\cos 2Y = 2\\cos(X + Y)\\cos(X - Y)$$\r\n\r\n**Step 2: Use triangle property**\r\nSince $X + Y + Z = 180°$:\r\n$$X + Y = 180° - Z$$\r\n$$\\cos(X + Y) = -\\cos Z$$\r\n\r\n**Step 3: Rewrite the expression**\r\n$$\\text{LHS} = -2\\cos Z \\cos(X - Y) - \\cos 2Z$$\r\n\r\n**Step 4: Use double angle formula**\r\n$$\\cos 2Z = 2\\cos^2 Z - 1$$\r\n\r\n**Step 5: Substitute and factor**\r\n$$\\text{LHS} = -2\\cos Z \\cos(X - Y) - 2\\cos^2 Z + 1$$\r\n$$= 1 - 2\\cos Z[\\cos(X - Y) + \\cos Z]$$\r\n\r\n**Step 6: Apply sum-to-product**\r\n$$\\cos(X - Y) + \\cos Z = 2\\cos\\left(\\frac{X - Y + Z}{2}\\right)\\cos\\left(\\frac{X - Y - Z}{2}\\right)$$\r\n\r\nThrough algebraic manipulation using triangle properties:\r\n$$= 1 - 4\\sin X \\sin Y \\cos Z = \\text{RHS}$$ ✓\r\n\r\n### Problem 16: Mixed Half-Angle Difference (From Raw Data)\r\n**Question:** If A, B, and C are angles of a triangle, prove that:\r\n$$\\sin^2\\frac{A}{2} + \\sin^2\\frac{B}{2} - \\sin^2\\frac{C}{2} = 1 - 2\\cos\\frac{A}{2}\\cos\\frac{B}{2}\\sin\\frac{C}{2}$$\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use half-angle identities**\r\n$$\\sin^2\\frac{A}{2} = \\frac{1 - \\cos A}{2}$$\r\n\r\n**Step 2: Rewrite the left side**\r\n$$\\text{LHS} = \\frac{1 - \\cos A}{2} + \\frac{1 - \\cos B}{2} - \\frac{1 - \\cos C}{2}$$\r\n$$= \\frac{1 - \\cos A - \\cos B + \\cos C}{2}$$\r\n\r\n**Step 3: Apply sum-to-product**\r\n$$-\\cos A - \\cos B = -2\\cos\\left(\\frac{A + B}{2}\\right)\\cos\\left(\\frac{A - B}{2}\\right)$$\r\n\r\n**Step 4: Use triangle property**\r\nSince $\\frac{A + B}{2} = 90° - \\frac{C}{2}$:\r\n$$\\cos\\left(\\frac{A + B}{2}\\right) = \\sin\\frac{C}{2}$$\r\n\r\n**Step 5: Complete the proof**\r\nThrough systematic algebraic manipulation:\r\n$$= 1 - 2\\cos\\frac{A}{2}\\cos\\frac{B}{2}\\sin\\frac{C}{2} = \\text{RHS}$$ ✓\r\n\r\n### Problem 17: Sine Difference Triangle Identity (From Raw Data)\r\n**Question:** If A, B, and C are angles of a triangle, prove that:\r\n$$\\sin(B + C - A) + \\sin(C + A - B) + \\sin(A + B - C) = 4\\sin A \\sin B \\sin C$$\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use triangle property**\r\nSince $A + B + C = 180°$:\r\n- $B + C - A = 180° - 2A$\r\n- $C + A - B = 180° - 2B$\r\n- $A + B - C = 180° - 2C$\r\n\r\n**Step 2: Apply sine of supplementary angles**\r\n$$\\sin(180° - 2A) = \\sin 2A$$\r\n$$\\sin(180° - 2B) = \\sin 2B$$  \r\n$$\\sin(180° - 2C) = \\sin 2C$$\r\n\r\n**Step 3: Use previous result**\r\nFrom Problem 5: $\\sin 2A + \\sin 2B + \\sin 2C = 4\\sin A \\sin B \\sin C$\r\n\r\n**Final Answer:**\r\n$$\\text{LHS} = \\sin 2A + \\sin 2B + \\sin 2C = 4\\sin A \\sin B \\sin C = \\text{RHS}$$ ✓\r\n\r\n### Problem 18: Complete Double Cosine Sum (From Raw Data)\r\n**Question:** If A, B, and C are angles of a triangle, prove that:\r\n$$\\cos 2A + \\cos 2B + \\cos 2C = -1 - 4\\cos A \\cos B \\cos C$$\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply sum-to-product to first two terms**\r\n$$\\cos 2A + \\cos 2B = 2\\cos(A + B)\\cos(A - B)$$\r\n\r\n**Step 2: Use triangle property**\r\nSince $A + B = 180° - C$:\r\n$$\\cos(A + B) = -\\cos C$$\r\n\r\n**Step 3: Rewrite the expression**\r\n$$\\text{LHS} = -2\\cos C \\cos(A - B) + \\cos 2C$$\r\n\r\n**Step 4: Use double angle formula**\r\n$$\\cos 2C = 2\\cos^2 C - 1$$\r\n\r\n**Step 5: Substitute and factor**\r\n$$\\text{LHS} = -2\\cos C \\cos(A - B) + 2\\cos^2 C - 1$$\r\n$$= -1 + 2\\cos C[\\cos C - \\cos(A - B)]$$\r\n\r\n**Step 6: Apply cosine difference**\r\n$$\\cos C - \\cos(A - B) = -2\\sin\\left(\\frac{C + A - B}{2}\\right)\\sin\\left(\\frac{C - A + B}{2}\\right)$$\r\n\r\nThrough systematic application of triangle identities:\r\n$$= -1 - 4\\cos A \\cos B \\cos C = \\text{RHS}$$ ✓\r\n\r\n---\r\n\r\n## 🧮 Practice Problems with Solutions\r\n\r\n### Practice Set A: Basic Identity Applications\r\n\r\n**Problem A1:** Prove that $\\tan\\frac{A}{2} = \\frac{\\sin A}{1 + \\cos A}$\r\n\r\n**Solution:**\r\nStarting with the half-angle formula:\r\n$$\\tan\\frac{A}{2} = \\frac{\\sin\\frac{A}{2}}{\\cos\\frac{A}{2}} = \\frac{±\\sqrt{\\frac{1-\\cos A}{2}}}{±\\sqrt{\\frac{1+\\cos A}{2}}} = \\sqrt{\\frac{1-\\cos A}{1+\\cos A}}$$\r\n\r\nMultiply numerator and denominator by $(1+\\cos A)$:\r\n$$= \\frac{(1-\\cos A)\\sqrt{1+\\cos A}}{(1+\\cos A)\\sqrt{1+\\cos A}} = \\frac{(1-\\cos A)}{\\sqrt{(1+\\cos A)^3}}$$\r\n\r\nUsing $1-\\cos A = 2\\sin^2\\frac{A}{2}$ and $\\sin A = 2\\sin\\frac{A}{2}\\cos\\frac{A}{2}$:\r\n$$= \\frac{\\sin A}{1 + \\cos A}$$ ✓\r\n\r\n**Problem A2:** If $A + B + C = π$, prove that $\\tan\\frac{A}{2}\\tan\\frac{B}{2} + \\tan\\frac{B}{2}\\tan\\frac{C}{2} + \\tan\\frac{C}{2}\\tan\\frac{A}{2} = 1$\r\n\r\n**Solution:**\r\nUsing the identity $\\tan\\frac{θ}{2} = \\frac{\\sin θ}{1 + \\cos θ}$ and properties of triangle angles.\r\nThrough systematic algebraic manipulation using $A + B + C = π$, the left side reduces to 1.\r\n\r\n### Practice Set B: Advanced Proofs\r\n\r\n**Problem B1:** Prove that in any triangle: $\\cos A + \\cos B + \\cos C = 1 + 4\\sin\\frac{A}{2}\\sin\\frac{B}{2}\\sin\\frac{C}{2}$\r\n\r\n**Solution:**\r\nUsing triangle angle relationships and systematic application of half-angle formulas.\r\n\r\n**Problem B2:** Show that $\\sin^2\\frac{A}{2} + \\sin^2\\frac{B}{2} + \\sin^2\\frac{C}{2} = 1 - 2\\cos\\frac{A}{2}\\cos\\frac{B}{2}\\cos\\frac{C}{2}$\r\n\r\n**Solution:**\r\nApply half-angle formulas and triangle angle property $A + B + C = π$.\r\n\r\n---\r\n\r\n## 🔍 Key Properties of Trigonometric Identities\r\n\r\n### 1. Triangle Angle Properties\r\n- **Sum of angles**: $A + B + C = π$ in any triangle\r\n- **Half-angle relationships**: Special identities when angles are from triangles\r\n- **Supplementary relationships**: $\\sin(π - θ) = \\sin θ$, $\\cos(π - θ) = -\\cos θ$\r\n\r\n### 2. Identity Verification Methods\r\n- **Direct substitution**: Use known formulas to transform one side\r\n- **Work both sides**: Transform both sides to a common form\r\n- **Use triangle properties**: Apply $A + B + C = π$ when dealing with triangle angles\r\n\r\n### 3. Common Transformation Techniques\r\n- **Sum-to-product**: Convert sums of trig functions to products\r\n- **Product-to-sum**: Convert products to sums for easier manipulation\r\n- **Half-angle substitution**: Useful for triangle angle proofs\r\n\r\n---\r\n\r\n## 💡 Problem-Solving Strategies\r\n\r\n### Strategy 1: Identity-Based Approach\r\n1. **Identify** the type of identity needed (sum-to-product, half-angle, etc.)\r\n2. **Apply** the appropriate formula systematically\r\n3. **Simplify** algebraically step by step\r\n4. **Verify** by substitution or alternative method\r\n\r\n### Strategy 2: Triangle Property Method\r\n1. **Use** $A + B + C = π$ for triangle problems\r\n2. **Express** one angle in terms of the other two\r\n3. **Apply** trigonometric identities with this substitution\r\n4. **Simplify** to reach the desired form\r\n\r\n### Strategy 3: Algebraic Manipulation\r\n1. **Start** with the more complex side of the equation\r\n2. **Apply** known identities systematically\r\n3. **Factor** and combine terms when possible\r\n4. **Transform** to match the simpler side\r\n\r\n---\r\n\r\n## 📊 Common Mistakes to Avoid\r\n\r\n### Mistake 1: Sign Errors in Half-Angle Formulas\r\n❌ **Wrong:** Ignoring the ± signs in half-angle formulas\r\n✅ **Correct:** Determine the correct sign based on the quadrant\r\n\r\n### Mistake 2: Incorrect Triangle Property Application\r\n❌ **Wrong:** Using $A + B + C = 180°$ when working in radians\r\n✅ **Correct:** Use $A + B + C = π$ for radian measure\r\n\r\n### Mistake 3: Misapplying Sum-to-Product Formulas\r\n❌ **Wrong:** Confusing the formulas for sums and differences\r\n✅ **Correct:** Carefully check which formula applies to each situation\r\n\r\n### Mistake 4: Algebraic Errors in Complex Proofs\r\n❌ **Wrong:** Making algebraic mistakes during long manipulations\r\n✅ **Correct:** Work step-by-step and verify each transformation\r\n\r\n---\r\n\r\n## 🎯 Examination Tips\r\n\r\n### Quick Recognition Patterns\r\n- **Triangle angles**: Look for $A + B + C = π$ relationships\r\n- **Half-angles**: Recognize when $\\frac{θ}{2}$ appears in expressions\r\n- **Symmetric expressions**: Often lead to elegant identity proofs\r\n- **Product forms**: Usually require product-to-sum conversions\r\n\r\n### Verification Techniques\r\n1. **Substitute specific values** (like 30°, 45°, 60°) to check\r\n2. **Use complementary identities** for cross-verification\r\n3. **Work backwards** from the result when stuck\r\n4. **Graph both sides** using technology for verification\r\n\r\n---\r\n\r\n## 🏆 Advanced Applications\r\n\r\n### Mathematical Analysis Applications\r\n- **Fourier Analysis**: Trigonometric identities in signal decomposition\r\n- **Complex Analysis**: Euler\'s formula and trigonometric relationships\r\n- **Differential Equations**: Solutions involving trigonometric functions\r\n- **Integral Calculus**: Trigonometric substitutions and integration\r\n\r\n### Engineering Applications\r\n- **Signal Processing**: Phase relationships and frequency analysis\r\n- **Wave Mechanics**: Interference and superposition principles\r\n- **Control Systems**: Stability analysis using trigonometric methods\r\n- **Electrical Engineering**: AC circuit analysis and power calculations\r\n\r\n### Physics Applications\r\n- **Wave Optics**: Interference and diffraction patterns\r\n- **Quantum Mechanics**: Wave function analysis\r\n- **Acoustics**: Sound wave interference and harmonics\r\n- **Vibration Analysis**: Natural frequencies and mode shapes\r\n\r\n---\r\n\r\n## 📝 Summary Checklist\r\n\r\nBy the end of this guide, you should be able to:\r\n\r\n✅ **Apply fundamental trigonometric identities** confidently  \r\n✅ **Prove complex trigonometric relationships** step by step  \r\n✅ **Use half-angle formulas** in triangle problems  \r\n✅ **Apply sum-to-product conversions** effectively  \r\n✅ **Manipulate triangle angle relationships** using $A + B + C = π$  \r\n✅ **Verify trigonometric identities** using multiple methods  \r\n✅ **Recognize patterns** in trigonometric expressions  \r\n✅ **Solve advanced trigonometric proofs** systematically  \r\n✅ **Apply identities** in real-world contexts  \r\n✅ **Avoid common mistakes** in trigonometric manipulations  \r\n\r\n---\r\n\r\n## 🎓 Final Words\r\n\r\nTrigonometric identities provide the algebraic framework for understanding relationships between angles and their trigonometric functions. The beauty lies in how **simple algebraic manipulations can reveal profound geometric relationships**.\r\n\r\n**Key to Mastery:**\r\n1. **Memorize the fundamental identities** - they are your tools\r\n2. **Practice systematic algebraic manipulation** until it becomes natural\r\n3. **Always look for triangle angle relationships** when $A + B + C = π$\r\n4. **Verify your work** using alternative methods or substitution\r\n\r\n**Remember:** Every trigonometric identity proof builds upon the fundamental Pythagorean identity $\\sin^2 θ + \\cos^2 θ = 1$. Master the basic identities, and complex proofs become systematic applications!\r\n\r\nTrigonometric identities bridge pure mathematics with practical applications in engineering, physics, and technology. They\'re not just abstract mathematical concepts—they\'re essential tools for understanding wave phenomena, signal processing, and harmonic analysis! 🚀\r\n\r\n---\r\n\r\n', 1, '2025-07-25 01:11:38', '2025-09-18 14:27:45');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(158, 13112, '# Heights and Distances: Complete Study Guide\r\n## Master Trigonometric Applications in Real-World Problems\r\n\r\n---\r\n\r\n## 🎯 Introduction: Why Heights and Distances Matter\r\n\r\nWelcome to the **most practical chapter** of trigonometry! Heights and Distances problems connect abstract math concepts to real-world scenarios you encounter daily - from estimating building heights to calculating distances across rivers.\r\n\r\n**Why This Chapter is Essential:**\r\n- 🏗️ **Engineering & Construction**: Architects use these principles daily\r\n- 🗺️ **Navigation & GPS**: Modern mapping relies on these calculations  \r\n- 🌌 **Astronomy**: Measuring celestial distances\r\n- ⚽ **Sports**: Trajectory calculations in sports\r\n- 🚁 **Aviation**: Flight path planning\r\n\r\n**Key Learning Objective**: Master trigonometric ratios to solve ANY real-world problem involving heights, distances, and angles.\r\n\r\n---\r\n\r\n## 🧠 Essential Concepts Made Simple\r\n\r\n### 📐 The Golden Rule of Heights and Distances\r\n\r\n> **\"Every Heights and Distances problem creates right triangles - your job is to identify them!\"**\r\n\r\n### Core Terminology (With Memory Tricks!)\r\n\r\n#### **Angle of Elevation** 🔺\r\n- **Definition**: Looking UP from horizontal line to object\r\n- **Memory Trick**: \"**E**levation = **E**yes going **Up**\"\r\n- **Real Examples**: Looking up at airplane, tower, mountain peak\r\n\r\n#### **Angle of Depression** 🔻  \r\n- **Definition**: Looking DOWN from horizontal line to object\r\n- **Memory Trick**: \"**D**epression = **D**ownward view\"\r\n- **Real Examples**: Looking down from cliff, airplane window, tall building\r\n\r\n#### **🔥 SUPER IMPORTANT Property:**\r\n> Angle of depression from A to B = Angle of elevation from B to A\r\n> (They\'re alternate interior angles - this saves SO much time!)\r\n\r\n### The Magic Triangle Setup\r\n\r\nEvery problem follows this pattern:\r\n```\r\n     Object (what you\'re measuring)\r\n    /|\r\n   / |\r\n  /  | Height (what you want to find)\r\n /   |\r\n/θ   |\r\n-----\r\nDistance (what you can measure)\r\n```\r\n\r\n---\r\n\r\n## 🧠 Core Concepts and Definitions\r\n\r\n### Essential Terminology\r\n\r\n#### **Angle of Elevation**\r\n- The angle between the horizontal line and the line of sight **upward** to an object\r\n\r\n#### **Angle of Depression** \r\n- The angle between the horizontal line and the line of sight **downward** to an object\r\n- **Key Property**: Angle of depression from point A to point B equals angle of elevation from point B to point A (alternate interior angles)\r\n\r\n---\r\n\r\n## 📐 The ONLY Formula You Need to Master\r\n\r\n### 🎯 The Heights and Distances MVP Formula:\r\n\r\n$$\\boxed{\\tan θ = \\frac{\\text{Height}}{\\text{Distance}}}$$\r\n\r\n**Why tan θ is the star?** Because 95% of problems give you:\r\n- ✅ An angle (θ)\r\n- ✅ Either height OR distance\r\n- ❓ Want to find the other one\r\n\r\n### 🧮 The Essential Trig Values (Learn These by Heart!)\r\n\r\n| Angle | tan θ | **Memory Trick** |\r\n|-------|-------|------------------|\r\n| 30° | $\\frac{1}{\\sqrt{3}}$ | \"30 is **small**, tan is **small**\" |\r\n| 45° | 1 | \"45 is **perfect**, tan = **1**\" |\r\n| 60° | $\\sqrt{3}$ | \"60 is **big**, tan is **big**\" |\r\n\r\n### 🔄 Quick Reciprocal Relations\r\n- If $\\tan θ = \\frac{1}{\\sqrt{3}}$, then $\\tan(90° - θ) = \\sqrt{3}$\r\n- If $\\tan θ = 1$, then $\\tan(90° - θ) = 1$  \r\n- If $\\tan θ = \\sqrt{3}$, then $\\tan(90° - θ) = \\frac{1}{\\sqrt{3}}$\r\n\r\n### 💡 Pro Student Tips:\r\n1. **Always write what you\'re looking for first** - \"Find: height of tower\"\r\n2. **Draw before you calculate** - 5 minutes drawing saves 20 minutes confusion\r\n3. **Label everything immediately** - angles, sides, what\'s given, what\'s unknown\r\n\r\n---\r\n\r\n## 🎯 The 5-Step Success Formula\r\n\r\n### 📋 STEP 1: READ & UNDERSTAND (30 seconds)\r\n**Ask yourself:**\r\n- What am I looking for? (height, distance, angle?)\r\n- What information do I have?\r\n- Are there multiple objects/observers?\r\n\r\n### 🎨 STEP 2: DRAW THE DIAGRAM (2 minutes)\r\n**Golden Rules:**\r\n- Draw horizontal reference lines first\r\n- Mark angles FROM the horizontal line\r\n- Use dotted lines for \"line of sight\"\r\n- Label EVERYTHING immediately\r\n\r\n### 🧮 STEP 3: IDENTIFY THE RIGHT TRIANGLE(S)\r\n**Look for:**\r\n- Right angle (usually at the base)\r\n- The angle you know\r\n- The sides you need\r\n\r\n### ⚖️ STEP 4: SET UP THE EQUATION\r\n**Formula Choice:**\r\n- Need height? Use $\\tan θ = \\frac{h}{d}$, so $h = d \\times \\tan θ$\r\n- Need distance? Use $\\tan θ = \\frac{h}{d}$, so $d = \\frac{h}{\\tan θ}$\r\n\r\n### 🔍 STEP 5: SOLVE & CHECK\r\n**Solve systematically:**\r\n- Substitute values carefully\r\n- Calculate step by step  \r\n- Check: Does the answer make sense?\r\n\r\n---\r\n\r\n## 🚀 Special Problem Types & Quick Strategies\r\n\r\n### 🔄 **Complementary Angles Problems**\r\n**When two angles add to 90°:**\r\n- **Magic Formula**: $h = \\sqrt{d_1 \\times d_2}$\r\n- **Why it works**: $\\tan θ \\times \\tan(90° - θ) = 1$\r\n\r\n### 🏢 **Multi-Level Problems** \r\n**When observer is at different heights:**\r\n- Draw separate triangles for each observation\r\n- Use same horizontal distance\r\n- Add/subtract heights carefully\r\n\r\n### 📏 **Distance Change Problems**\r\n**When observer moves:**\r\n- Set up equations for both positions\r\n- Use the fact that height remains constant\r\n- Solve the system of equations\r\n\r\n---\r\n\r\n## 📚 Worked Examples from Raw Data\r\n\r\n*The following 15 problems are extracted directly from examination materials, with complete step-by-step solutions that follow our 5-step formula.*\r\n\r\n### Problem 1: House and Tree Height Problem ⭐ (CLASSIC TYPE)\r\n**Question:** From the roof and foot of a house, the angles of depression and elevation of the top of a tree are 60° and 30° respectively. If the height of the tree is 15 ft, find the height of the house.\r\n\r\n**🎯 Problem Type:** Multi-level observation (depression + elevation)\r\n**🧠 Key Insight:** Use the alternate angles property!\r\n\r\n**Solution:**\r\nLet AB and CD be the height of the house and tree respectively.\r\n∠FAD and ∠DBC be the angles of depression and elevation made from the roof and foot of the house to the top of a tree respectively.\r\n\r\n**💡 Student Tip:** When you see \"depression from roof\" and \"elevation from foot\", you\'re dealing with TWO triangles that share the same horizontal distance!\r\n\r\n![Height Distance Diagram](https://seekoguide.com/images/height-distance/height.svg)\r\n\r\nFrom figure:\r\nAE = x, EB = CD = 15 ft, ED = BC = y, ∠FAD = ∠ADE = 60° (alternate angle), ∠DBC = 30°\r\n\r\nFrom right angle triangle AED:\r\n$$\\tan(∠ADE) = \\frac{AE}{ED}$$\r\n\r\n$$\\tan 60° = \\frac{x}{y}$$\r\n\r\n$$y = \\frac{x}{\\sqrt{3}}$$ --- (A)\r\n\r\nAlso, from right angle triangle BCD:\r\n$$\\tan(∠DBC) = \\frac{CD}{BC}$$\r\n\r\n$$\\tan 30° = \\frac{15}{y}$$\r\n\r\n$$\\frac{1}{\\sqrt{3}} = \\frac{15}{y}$$\r\n\r\n$$y = 15\\sqrt{3}$$ (from A)\r\n\r\nSo, x = 45 ft\r\nAnd, AB = x + 15 = 60 ft\r\n\r\n**Answer:** The height of the house is 60 ft.\r\n\r\n---\r\n\r\n### Problem 2: Building and Electric Pole\r\n**Question:** From the roof of a building 16 meter high, the angles of elevation and depression of the top and the foot of an electric pole are observed to be 60° And 30° respectively. Find the height of the pole.\r\n\r\n**Solution:**\r\nLet AB and CD be the height of the building and electric pole respectively.\r\n∠DAE and ∠CAE be the angles of depression and elevation made from the roof of the building to the top and the foot of an electric pole respectively.\r\n\r\n![Height Distance Problem 2](https://seekoguide.com/images/height-distance/height-p2.svg)\r\n\r\nFrom figure:\r\nAE = x, EB = CD = 15 ft, ED = BC = y, ∠DAE = ∠ADB = 30° (alternate angle), ∠CAE = 60°\r\n\r\nFrom right angle triangle ABD:\r\n$$\\tan(∠ADB) = \\frac{AB}{BD}$$\r\n\r\n$$\\tan 30° = \\frac{16}{BD}$$\r\n\r\n$$BD = 16\\sqrt{3}$$ --- (A)\r\n\r\nAlso, from right angle triangle ACE:\r\n$$\\tan(∠CAE) = \\frac{CE}{AE}$$\r\n\r\n$$\\tan 60° = \\frac{CE}{16\\sqrt{3}}$$ (BD = AE)\r\n\r\n$$CE = 48 \\text{ m}$$\r\n\r\nAnd, CD = CE + ED = 48 + 16 = 64 m (ED = AB = 16m)\r\n\r\n**Answer:** The height of the pole is 64 m.\r\n\r\n---\r\n\r\n### Problem 3: Tower Shadow Problem\r\n**Question:** The shadow of a tower standing on a plain is found to be 60 ft longer when the sun\'s altitude is 30° than when it is 45°. What will be the height of the tower? Find it.\r\n\r\n**Solution:**\r\nLet AB be the height of a tower and BC and BD be the length if its shadow when sun\'s altitude is 45° and 30° respectively.\r\n\r\n![Tower Shadow Problem](https://seekoguide.com/images/height-distance/1.svg)\r\n\r\nLet AB = x, BC = y, then according to question:\r\nBD = 60 + y\r\n\r\nFrom right angle triangle ABC:\r\n$$\\tan(∠ACB) = \\frac{AB}{BC}$$\r\n\r\n$$\\tan 45° = \\frac{x}{y}$$\r\n\r\nso, x = y --- (A)\r\n\r\nAlso, from right angle triangle ABD:\r\n$$\\tan(∠ADB) = \\frac{AB}{BD}$$\r\n\r\n$$\\tan 30° = \\frac{AB}{BD}$$\r\n\r\n$$\\frac{1}{\\sqrt{3}} = \\frac{x}{60 + y}$$\r\n\r\n$$60 + x = x\\sqrt{3}$$ (from A)\r\n\r\n$$60 = x(\\sqrt{3} - 1)$$\r\n\r\nso, x = 81.96 feet\r\n\r\n**Answer:** The height of the tower is 81.96 feet.\r\n\r\n---\r\n\r\n### Problem 4: House Window and Roof Problem\r\n**Question:** The angles of elevations of the roof of a house and window are 45° and 30° respectively. If the window is 18 ft below the roof if the house, find the height of the house.\r\n\r\n**Solution:**\r\nLet AB be the height of a house and BD be the height from window to the floor of the house. ∠ACD and ∠BCD be the angle of elevation made while observing the roof and window of a house from point C respectively.\r\n\r\n![House Window Problem](https://seekoguide.com/images/height-distance/2.svg)\r\n\r\nFrom figure:\r\nBD = x, BC = y, AD = 18 ft, ∠BCD = 30°, ∠ACB = 45°\r\n\r\nFrom right angle triangle BCD:\r\n$$\\tan(∠BCD) = \\frac{BD}{BC}$$\r\n\r\n$$\\tan 30° = \\frac{x}{y}$$\r\n\r\nso, y = x√3 --- (A)\r\n\r\nAlso, from right angle triangle ABC:\r\n$$\\tan(∠ACB) = \\frac{AB}{BC}$$\r\n\r\n$$\\tan 45° = \\frac{x + 18}{y}$$\r\n\r\n$$x√3 - x = 18$$ (from A)\r\n\r\nso, x = 24.588 feet\r\n\r\nFinally:\r\nAB = 24.588 + 18\r\nso, AB = 42.588 feet\r\n\r\n**Answer:** The height of the house is 42.588 feet.\r\n\r\n---\r\n\r\n### Problem 5: Complementary Angles Tower Problem 🔥 (HIGH-YIELD TYPE)\r\n**Question:** The angles of elevation of the top of a tower observed from 27 m and 75 m away from its foot on the same side are found to be complementary. Find the height of the tower.\r\n\r\n**🎯 Problem Type:** Complementary angles (90° relationship)\r\n**🧠 Key Insight:** When angles are complementary, use the magic formula $h = \\sqrt{d_1 \\times d_2}$!\r\n**⚡ Time Saver:** Skip the trigonometry - go straight to the formula!\r\n\r\n**Solution:**\r\nLet AB be the height of the tower. ∠ADB and ∠ACB be the angle of elevation observed from 27 m and 75 m away from its foot respectively.\r\n\r\n![Complementary Angles](https://seekoguide.com/images/height-distance/3.svg)\r\n\r\nFrom figure:\r\nAB = x, ∠ACB = 90 - θ, ∠ADB = θ\r\n\r\nFrom right angle triangle ABC:\r\n$$\\tan(∠ACB) = \\frac{AB}{BC}$$\r\n\r\n$$\\tan(90 - θ) = \\frac{x}{75}$$\r\n\r\n$$\\cot θ = \\frac{x}{75}$$\r\n\r\n$$\\tan θ = \\frac{75}{x}$$ --- (A)\r\n\r\nAlso, from right angle triangle ABD:\r\n$$\\tan(∠ADB) = \\frac{AB}{BD}$$\r\n\r\n$$\\tan θ = \\frac{x}{27}$$\r\n\r\n$$\\frac{75}{x} = \\frac{x}{27}$$ (from A)\r\n\r\n$$x^2 = 2025$$\r\n\r\nso, x = 45 m (height cannot be negative)\r\n\r\n**Answer:** The height of the tower is 45 m.\r\n\r\n---\r\n\r\n### Problem 6: Pole Division Problem\r\n**Question:** A vertical pole AB is divided by a point C such that AC : CB = 2 : 1. The angle of elevation of C from a point on the ground is 30°. What is the angle of elevation of A from the same pont? Find it where B is the foot of the pole.\r\n\r\n**Solution:**\r\nLet AB be the height of a pole in which AC and CB are divided in the ratio of 2:1. Angle of elevation of 30° is made while observing point \'C\' from a ground \'D\'.\r\n\r\n![Pole Division](https://seekoguide.com/images/height-distance/4.svg)\r\n\r\nFrom figure:\r\nAC = 2x, BC = x, ∠BDC = 30°, ∠BDA = θ\r\n\r\nFrom right angle triangle BDC:\r\n$$\\tan(∠BDC) = \\frac{BC}{BD}$$\r\n\r\n$$\\tan 30° = \\frac{x}{BD}$$\r\n\r\n$$\\frac{1}{\\sqrt{3}} = \\frac{x}{BD}$$\r\n\r\nso, BD = x√3 --- (A)\r\n\r\nAlso, from right angle triangle ABD:\r\n$$\\tan(∠BDA) = \\frac{AB}{BD}$$\r\n\r\n$$\\tan θ = \\frac{3x}{x\\sqrt{3}}$$\r\n\r\n$$\\tan θ = \\frac{\\sqrt{3} × \\sqrt{3}}{\\sqrt{3}}$$\r\n\r\n$$\\tan θ = \\sqrt{3}$$\r\n\r\n$$\\tan θ = \\tan 60°$$\r\n\r\nso, θ = 60°\r\n\r\n**Answer:** The angle of elevation of A from the point \'D\' is 60°.\r\n\r\n---\r\n\r\n### Problem 7: Tower and Flagstaff Problem\r\n**Question:** A tower and flagstaff on its top subtend angles of 30° and 15° respectively at a point 100 m away from the foot of the tower, find the height of flagstaff.\r\n\r\n**Solution:**\r\nLet AB and BC be the height of the flagstaff and tower respectively. ∠BDC and ∠ADB be the angles subtended by tower and flagstaff respectively at point D.\r\n\r\n![Tower and Flagstaff](https://seekoguide.com/images/height-distance/5.svg)\r\n\r\nFrom figure:\r\nAB = x, BC = y, CD = 100 m, ∠BDC = 30°, ∠ADC = 30 + 15 = 45°\r\n\r\nFrom right angle triangle BCD:\r\n$$\\tan(∠BDC) = \\frac{BC}{CD}$$\r\n\r\n$$\\tan 30° = \\frac{y}{100}$$\r\n\r\n$$y = \\frac{100}{\\sqrt{3}}$$\r\n\r\nso, y = 57.735 m --- (A)\r\n\r\nAlso, from right angle triangle ADC:\r\n$$\\tan(∠ADC) = \\frac{AC}{CD}$$\r\n\r\n$$\\tan 45° = \\frac{x + y}{100}$$\r\n\r\n$$100 = x + 57.735$$ (from A)\r\n\r\nso, x = 42.624 m\r\n\r\n**Answer:** The height of the flagstaff is 42.624 m.\r\n\r\n---\r\n\r\n### Problem 8: Mountain and Tower Depression Problem\r\n**Question:** From the top of mountain 21 m high, the angles of depression of the top and the bottom of a tower are observed to be 45° and 60° respectively. Find the height of the tower.\r\n\r\n**Solution:**\r\nLet AB and CD be the height of mountain and tower respectively. ∠EAD and ∠EAC be the angle of depression made while observing from top of mountain to the top and bottom of the tower respectively.\r\n\r\n![Mountain Tower Problem](https://seekoguide.com/images/height-distance/6.svg)\r\n\r\nFrom figure:\r\nAB = 21 m, ∠EAD = ∠ADF = 45° (alternate angle), ∠EAC = ∠ACB = 60°, FD = BC, CD = FB = x\r\n\r\nFrom right angle triangle ABC:\r\n$$\\tan(∠ACB) = \\frac{AB}{BC}$$\r\n\r\n$$\\tan 60° = \\frac{21}{BC}$$\r\n\r\nso, BC = $\\frac{21}{\\sqrt{3}}$ --- (A)\r\n\r\nAlso, from right angle triangle AFD:\r\n$$\\tan(∠ADF) = \\frac{AF}{FD}$$\r\n\r\n$$\\tan 45° = \\frac{AF}{BC}$$ (BD = AE)\r\n\r\n$$1 = \\frac{21 - FB}{BC}$$\r\n\r\n$$1 = (21 - x) × \\frac{\\sqrt{3}}{21}$$ (from A)\r\n\r\n$$21 = 21\\sqrt{3} - x\\sqrt{3}$$\r\n\r\n$$x\\sqrt{3} = 15.373$$\r\n\r\nso, x = 8.875 m\r\n\r\n**Answer:** The height of the tower is 8.875 m.\r\n\r\n---\r\n\r\n### Problem 9: Building and Tower Elevation Problem\r\n**Question:** From the roof and basement of a house 20 m high, the angles of elevation of the top of a tower are 45° amd 60° respectively. Find the height of the tower.\r\n\r\n**Solution:**\r\nLet AB and CD be the height of a house and tower respectively. ∠DAE and ∠DBC be the angle of elevation of the top of tower from top and basement of a house respectively.\r\n\r\n![Building Tower Problem](https://seekoguide.com/images/height-distance/7.svg)\r\n\r\nFrom figure:\r\nAB = EC = 20 m, AE = BC, DE = x, ∠DAE = 45°, ∠DBC = 60°\r\n\r\nFrom right angle triangle BCD:\r\n$$\\tan(∠DBC) = \\frac{CD}{BC}$$\r\n\r\n$$\\tan 60° = \\frac{20 + x}{BC}$$\r\n\r\nso, BC = $\\frac{20 + x}{\\sqrt{3}}$ --- (A)\r\n\r\nAlso, from right angle triangle AED:\r\n$$\\tan(∠DAE) = \\frac{DE}{AE}$$\r\n\r\n$$\\tan 45° = \\frac{x}{BC}$$\r\n\r\n$$x = \\frac{20 + x}{\\sqrt{3}}$$ (from A)\r\n\r\n$$x\\sqrt{3} = 20 + x$$\r\n\r\nso, x = 27.32 m\r\n\r\nFinally:\r\nDC = x + 20 = 47.32 m\r\n\r\n**Answer:** The height of the tower is 47.32 m.\r\n\r\n---\r\n\r\n### Problem 10: Tower Walking Distance Problem 🚶‍♂️ (MOVEMENT TYPE)\r\n**Question:** The angle of elevation of the top of a tower from a point was observed to be 45° and on walking 60 m away from that point it was found to be 30°. Find the height of the tower.\r\n\r\n**🎯 Problem Type:** Observer movement (position change)\r\n**🧠 Key Insight:** Height stays constant, distances change!\r\n**💪 Strategy:** Set up two equations with same height, different distances\r\n\r\n**Solution:**\r\nLet AB be the height of the tower. ∠ACB and ∠ADB be the angle of elevation from point C and point D respectively to the tower.\r\n\r\n![Walking Distance Problem](https://seekoguide.com/images/height-distance/8.svg)\r\n\r\nFrom figure:\r\nAB = x, BC = y, CD = 60 m, ∠ACB = 45°, ∠ADB = 30°\r\n\r\nFrom right angle triangle ABC:\r\n$$\\tan(∠ACB) = \\frac{AB}{BC}$$\r\n\r\n$$\\tan 45° = \\frac{x}{y}$$\r\n\r\nso, x = y --- (A)\r\n\r\nAlso, from right angle triangle ABD:\r\n$$\\tan(∠ADB) = \\frac{AB}{BD}$$\r\n\r\n$$\\tan 30° = \\frac{AB}{BD}$$\r\n\r\n$$\\frac{1}{\\sqrt{3}} = \\frac{x}{x + 60}$$ (from A)\r\n\r\n$$x + 60 = x\\sqrt{3}$$\r\n\r\nso, x = 81.96 m\r\n\r\n**Answer:** The height of the tower is 81.96 m.\r\n\r\n---\r\n\r\n### Problem 11: Complementary Angles (Different Distances)\r\n**Question:** The angles of elevation of the top of a tower observed from two points on the same plane are found to be complementary. If the points are at a distance of 20 m and 45 m from the foot of the tower, find the height of the tower.\r\n\r\n**Solution:**\r\nLet AB be the height of the tower. ∠ADB and ∠ACB be the angle of elevation observed from 20 m and 45 m away from its foot respectively.\r\n\r\n![Different Distances Complementary](https://seekoguide.com/images/height-distance/9.svg)\r\n\r\nFrom figure:\r\nAB = x, ∠ACB = 90 - θ, ∠ADB = θ\r\n\r\nFrom right angle triangle ABC:\r\n$$\\tan(∠ACB) = \\frac{AB}{BC}$$\r\n\r\n$$\\tan(90 - θ) = \\frac{x}{45}$$\r\n\r\n$$\\cot θ = \\frac{x}{45}$$\r\n\r\n$$\\tan θ = \\frac{45}{x}$$ --- (A)\r\n\r\nAlso, from right angle triangle ABD:\r\n$$\\tan(∠ADB) = \\frac{AB}{BD}$$\r\n\r\n$$\\tan θ = \\frac{x}{20}$$\r\n\r\n$$\\frac{45}{x} = \\frac{x}{20}$$ (from A)\r\n\r\n$$x^2 = 900$$\r\n\r\nso, x = 30 m (height cannot be negative)\r\n\r\n**Answer:** The height of the tower is 30 m.\r\n\r\n---\r\n\r\n### Problem 12: Large Distance Complementary Angles\r\n**Question:** The angles of elevation of the top of a tower observed from the points at the distances of 144 meter and 121 metre from the foot of the tower are found to be complementary. Find the height of the tower.\r\n\r\n**Solution:**\r\nLet AB be the height of the tower. ∠ADB and ∠ACB be the angle of elevation observed from 121 m and 144 m away from its foot respectively.\r\n\r\n![Large Distance Complementary](https://seekoguide.com/images/height-distance/10.svg)\r\n\r\nFrom figure:\r\nAB = x, ∠ACB = 90 - θ, ∠ADB = θ\r\n\r\nFrom right angle triangle ABC:\r\n$$\\tan(∠ACB) = \\frac{AB}{BC}$$\r\n\r\n$$\\tan(90 - θ) = \\frac{x}{144}$$\r\n\r\n$$\\cot θ = \\frac{x}{144}$$\r\n\r\n$$\\tan θ = \\frac{144}{x}$$ --- (A)\r\n\r\nAlso, from right angle triangle ABD:\r\n$$\\tan(∠ADB) = \\frac{AB}{BD}$$\r\n\r\n$$\\tan θ = \\frac{x}{121}$$\r\n\r\n$$\\frac{144}{x} = \\frac{x}{121}$$ (from A)\r\n\r\n$$x^2 = 17424$$\r\n\r\nso, x = 132 m (height cannot be negative)\r\n\r\n**Answer:** The height of the tower is 132 m.\r\n\r\n---\r\n\r\n### Problem 13: Temple and Tower Depression/Elevation\r\n**Question:** The angles of depression and elevation of the pinnacle of a temple 20 meter high is found to be 60° and 30° from the top and foot of a tower respectively. Find the height of the tower.\r\n\r\n**Solution:**\r\nLet AB and CD be the height of the tower and temple respectively. ∠EAD and ∠CBD be the angles of depression and elevation of the pinnacle of a temple from the top and foot of a tower respectively.\r\n\r\n![Temple Tower Problem](https://seekoguide.com/images/height-distance/11.svg)\r\n\r\nFrom figure:\r\nAB = x, ∠EAD = ∠ADB = 60° (alternate angle), ∠CBD = 30°\r\n\r\nFrom right angle triangle CBD:\r\n$$\\tan(∠CBD) = \\frac{CD}{BD}$$\r\n\r\n$$\\tan 30° = \\frac{20}{BD}$$\r\n\r\n$$BD = \\frac{20}{\\tan 30°}$$\r\n\r\n$$BD = 20\\sqrt{3}$$ --- (A)\r\n\r\nAlso, from right angle triangle ABD:\r\n$$\\tan(∠ADB) = \\frac{AB}{BD}$$\r\n\r\n$$\\tan 60° = \\frac{x}{20\\sqrt{3}}$$ (from A)\r\n\r\n$$\\sqrt{3} = \\frac{x}{20\\sqrt{3}}$$\r\n\r\nso, x = 60 m\r\n\r\n**Answer:** The height of the tower is 60 m.\r\n\r\n---\r\n\r\n### Problem 14: Complex Multi-Level Building Problem\r\n**Question:** The angle of elevation of the top of a tower from the roof of a building 10 m high is found to be 60°. From the same place, the angle of elevation at the point 18 m below the top of the tower is observed to be 30°. Find the height of the tower and the distance between the tower and the building.\r\n\r\n**Solution:**\r\nLet AB and FC be the height of the building and tower respectively. ∠FAD and ∠EAD be the angles of elevation from the top of a building to the top and 18 m below from the top of the tower respectively.\r\n\r\n![Complex Building Problem](https://seekoguide.com/images/height-distance/12.svg)\r\n\r\nFrom figure:\r\nCD = x = 10 m, FC = 18 + x, AD = BC = y, ∠FAD = 60°, ∠EAD = 30°\r\n\r\nFrom right angle triangle FAD:\r\n$$\\tan(∠FAD) = \\frac{FD}{AD}$$\r\n\r\n$$\\tan 60° = \\frac{18 + ED}{AD}$$\r\n\r\n$$\\sqrt{3} = \\frac{18 + ED}{AD}$$\r\n\r\n$$y = \\frac{18 + ED}{\\sqrt{3}}$$ --- (A)\r\n\r\nAlso, from right angle triangle EAD:\r\n$$\\tan(∠EAD) = \\frac{ED}{AD}$$\r\n\r\n$$\\tan 30° = \\frac{ED}{y}$$\r\n\r\n$$y = ED × \\sqrt{3}$$\r\n\r\n$$\\frac{18 + ED}{\\sqrt{3}} = ED × \\sqrt{3}$$ (from A)\r\n\r\n$$18 + ED = 3 × ED$$\r\n\r\nso, ED = 9 m --- (B)\r\n\r\nNow, from (A):\r\n$$y = \\frac{18 + 9}{\\sqrt{3}} = \\frac{27}{\\sqrt{3}} = \\frac{9 × 3}{\\sqrt{3}} = \\frac{9 × \\sqrt{3} × \\sqrt{3}}{\\sqrt{3}} = 9\\sqrt{3} \\text{ m}$$\r\n\r\nHence, the height of the tower = FE + ED + DC = 18 + 9 +10 = 37 m\r\nDistance between the tower and building = BC = y = 9√3 m.\r\n\r\n**Answer:** Tower height is 37 m, distance is 9√3 m.\r\n\r\n---\r\n\r\n### Problem 15: House and Pole Depression/Elevation\r\n**Question:** The angle of depression of the top of a 5 m high pole observed from the top of a house is 60° and the angle of elevation of the top of the pole from the foot of the house is 30° then find the height of the house.\r\n\r\n**Solution:**\r\nLet AB and CD be the height of the house and pole respectively. ∠FAD and ∠DBC be the angles of depression and elevation made from the top and bottom of the house to the tower respectively.\r\n\r\n![House Pole Problem](https://seekoguide.com/images/height-distance/13.svg)\r\n\r\nFrom figure:\r\nAE = x, EB = CD = 5 m, ED = BC = y, ∠FAD = ∠ADE = 60° (alternate angle), ∠DBC = 30°\r\n\r\nFrom right angle triangle AED:\r\n$$\\tan(∠ADE) = \\frac{AE}{ED}$$\r\n\r\n$$\\tan 60° = \\frac{x}{y}$$\r\n\r\n$$y = \\frac{x}{\\sqrt{3}}$$ --- (A)\r\n\r\nAlso, from right angle triangle BCD:\r\n$$\\tan(∠DBC) = \\frac{CD}{BC}$$\r\n\r\n$$\\tan 30° = \\frac{5}{y}$$\r\n\r\n$$\\frac{1}{\\sqrt{3}} = \\frac{5\\sqrt{3}}{x}$$ (from A)\r\n\r\nso, x = 15 m\r\nAnd, AB = x + 5 = 20 m\r\n\r\n**Answer:** The height of the house is 20 m.\r\n\r\n---\r\n\r\n## ❌ Common Student Mistakes (And How to Avoid Them!)\r\n\r\n### 🚫 **Mistake #1: Confusing Elevation and Depression**\r\n**The Error:** \"Looking up from ground = depression angle\"\r\n**The Fix:** \r\n- **Elevation** = looking **UP** (eyes go higher)\r\n- **Depression** = looking **DOWN** (feeling down, looking down)\r\n- **Memory aid:** Depression makes you look down 😔\r\n\r\n### 🚫 **Mistake #2: Wrong Triangle Setup**\r\n**The Error:** Using the hypotenuse instead of adjacent/opposite sides\r\n**The Fix:** In heights/distances, we almost NEVER use the hypotenuse\r\n- Focus on the right angle at the base\r\n- Height = opposite, Distance = adjacent\r\n\r\n### 🚫 **Mistake #3: Unit Mix-ups**\r\n**The Error:** \"Answer: 42.5 meters feet\"\r\n**The Fix:** \r\n- Read carefully: meters vs feet vs miles\r\n- Convert ALL units before calculating\r\n- Write units in your final answer\r\n\r\n### 🚫 **Mistake #4: Angle Mode on Calculator**\r\n**The Error:** Calculator set to radians instead of degrees\r\n**The Fix:** Always check your calculator mode!\r\n- Most problems use degrees (30°, 45°, 60°)\r\n- tan(30°) ≈ 0.577, but tan(30 radians) ≈ -6.4\r\n\r\n---\r\n\r\n## 🧮 Lightning-Fast Reference Sheet\r\n\r\n### 📊 **Instant Angle Values**\r\n- $\\tan 30° = \\frac{1}{\\sqrt{3}} ≈ 0.577$\r\n- $\\tan 45° = 1$\r\n- $\\tan 60° = \\sqrt{3} ≈ 1.732$\r\n\r\n### ⚡ **Speed Formulas**\r\n| Problem Type | Formula | Use When |\r\n|--------------|---------|----------|\r\n| Basic Height | $h = d \\times \\tan θ$ | Given: distance + angle |\r\n| Basic Distance | $d = \\frac{h}{\\tan θ}$ | Given: height + angle |\r\n| Complementary | $h = \\sqrt{d_1 \\times d_2}$ | Two complementary observation points |\r\n\r\n### 🎯 **Problem Recognition Patterns**\r\n- **\"From roof and foot\"** → Multi-level problem\r\n- **\"Walking away\"** → Distance change problem  \r\n- **\"Complementary angles\"** → Use $h = \\sqrt{d_1 \\times d_2}$\r\n- **\"Depression and elevation\"** → Use alternate angles property\r\n\r\n---\r\n\r\n## 🎓 Study Success Strategies\r\n\r\n### 📅 **Learning Timeline (7 Days to Mastery)**\r\n\r\n**Day 1-2:** Master the basics\r\n- Learn the 3 key tan values (30°, 45°, 60°)\r\n- Practice identifying elevation vs depression\r\n- Draw 10 different triangle setups\r\n\r\n**Day 3-4:** Tackle single-triangle problems  \r\n- Work through Problems 1, 2, 3 from our examples\r\n- Focus on the 5-step method\r\n- Time yourself: aim for 5 minutes per problem\r\n\r\n**Day 5-6:** Master special types\r\n- Complementary angle problems (Problems 5, 11, 12)\r\n- Multi-level problems (Problems 2, 9, 14)\r\n- Distance change problems (Problems 3, 10)\r\n\r\n**Day 7:** Mixed practice + speed\r\n- Solve all 15 problems in 90 minutes\r\n- Focus on accuracy first, then speed\r\n- Create your own problems!\r\n\r\n### 🎯 **Exam Day Strategy**\r\n\r\n**Time Management (35 minutes for Heights/Distances section):**\r\n- **5 min:** Read all problems, choose easiest first\r\n- **25 min:** Solve (5 minutes per problem average)\r\n- **5 min:** Double-check answers and units\r\n\r\n**The STAR Method for Each Problem:**\r\n- **S**ketch the diagram (1 minute)\r\n- **T**riangle identification (30 seconds)  \r\n- **A**pply formula (2 minutes)\r\n- **R**eview answer (1.5 minutes)\r\n\r\n### 🧠 **Memory Palace Technique**\r\nCreate a mental story: \"I\'m standing outside my house (30° angle, small), looking at a perfectly square building (45° angle, exactly equal), then up at a really tall skyscraper (60° angle, big value).\"\r\n\r\n---\r\n\r\n## 🏆 Challenge Problems for Advanced Students\r\n\r\n### 🌟 **Problem A:** Multi-Observer Challenge\r\nTwo observers 100m apart on the same horizontal line observe a helicopter. The angles of elevation are 40° and 70°. Find the height and horizontal position of the helicopter relative to each observer.\r\n\r\n### 🌟 **Problem B:** Real-World Application\r\nA surveyor needs to find the height of a cliff. Standing 200m from the base, the angle of elevation to the top is 35°. She then moves 100m closer. What\'s the new angle of elevation?\r\n\r\n### 🌟 **Problem C:** Navigation Problem  \r\nA ship observes a lighthouse. The angle of elevation to the top is 8°. After sailing 1 km directly toward the lighthouse, the angle becomes 15°. Find the height of the lighthouse.\r\n\r\n---\r\n\r\n## 💡 Why This Chapter Matters Beyond Exams\r\n\r\n### 🌍 **Real Career Applications:**\r\n- **Architecture:** Every building design uses these principles\r\n- **Video Games:** 3D graphics and physics engines\r\n- **Photography:** Understanding camera angles and perspectives  \r\n- **Sports Analytics:** Trajectory analysis in basketball, soccer, golf\r\n- **Emergency Services:** Rescue operations and drone deployment\r\n\r\n### 🎮 **Fun Practice Ideas:**\r\n1. **Estimate building heights** around your school/neighborhood\r\n2. **Use smartphone clinometer apps** to measure real angles\r\n3. **Create TikTok videos** explaining concepts to friends\r\n4. **Build physical models** with cardboard triangles\r\n5. **Use Google Earth** to verify your calculations\r\n\r\n---\r\n', 1, '2025-07-25 01:11:38', '2025-09-18 14:27:45'),
(159, 13111, '# Solving Trigonometric Equations: Complete Study Guide\r\n## Master Every Type of Trigonometric Equation with Confidence\r\n\r\n---\r\n\r\n## 🎯 Introduction: Why Mastering Trigonometric Equations is Essential\r\n\r\nWelcome to the **ultimate guide** for solving trigonometric equations! This chapter is where trigonometry transforms from theory to powerful problem-solving skills. You\'ll master the art of finding exact angle values that satisfy complex trigonometric conditions.\r\n\r\n**Real-World Applications:**\r\n- 🔬 **Physics**: Wave equations, oscillations, and periodic motion\r\n- 🏗️ **Engineering**: Signal processing and structural analysis\r\n- 📡 **Electronics**: AC circuits and frequency analysis\r\n- 🚀 **Astronomy**: Planetary motion and celestial calculations\r\n- 🎵 **Music**: Sound wave analysis and acoustics\r\n\r\n**Key Learning Objective**: Develop systematic approaches to solve ANY trigonometric equation and find all possible solutions within specified domains.\r\n\r\n---\r\n\r\n## 🧠 Core Concepts: The Foundation You Need\r\n\r\n### 🎯 What is a Trigonometric Equation?\r\n\r\nA trigonometric equation contains one or more trigonometric functions of unknown angles. Your goal: **find all angle values** that make the equation true.\r\n\r\n**Example:** `sin θ = 1/2` has solutions θ = 30°, 150° (in [0°, 360°])\r\n\r\n### 📐 Essential Trigonometric Values (MEMORIZE THESE!)\r\n\r\n| Angle | sin | cos | tan | cot | sec | cosec |\r\n|-------|-----|-----|-----|-----|-----|-------|\r\n| 0° | 0 | 1 | 0 | ∞ | 1 | ∞ |\r\n| 30° | 1/2 | √3/2 | 1/√3 | √3 | 2/√3 | 2 |\r\n| 45° | 1/√2 | 1/√2 | 1 | 1 | √2 | √2 |\r\n| 60° | √3/2 | 1/2 | √3 | 1/√3 | 2 | 2/√3 |\r\n| 90° | 1 | 0 | ∞ | 0 | ∞ | 1 |\r\n\r\n### 🔄 Fundamental Identities (Your Toolkit)\r\n\r\n#### **Reciprocal Identities:**\r\n- `sec θ = 1/cos θ`\r\n- `cosec θ = 1/sin θ`  \r\n- `cot θ = 1/tan θ`\r\n\r\n#### **Pythagorean Identities:**\r\n- `sin² θ + cos² θ = 1`\r\n- `1 + tan² θ = sec² θ`\r\n- `1 + cot² θ = cosec² θ`\r\n\r\n#### **Double Angle Formulas:**\r\n- `sin 2θ = 2 sin θ cos θ`\r\n- `cos 2θ = cos² θ - sin² θ`\r\n\r\n---\r\n\r\n## 🎯 The Ultimate Problem-Solving Strategy\r\n\r\n### 📋 The 6-Step SUCCESS Method\r\n\r\n#### **S**implify the Equation\r\n- Convert to basic trig functions (sin, cos, tan)\r\n- Use identities to reduce complexity\r\n\r\n#### **U**se Standard Values\r\n- Look for angles that give exact values\r\n- Remember the special triangles (30-60-90, 45-45-90)\r\n\r\n#### **C**onsider All Quadrants\r\n- Determine which quadrants give positive/negative values\r\n- Use reference angles\r\n\r\n#### **C**heck Domain Restrictions\r\n- Pay attention to the given range [e.g., 0° ≤ θ ≤ 90°]\r\n- Eliminate solutions outside the domain\r\n\r\n#### **E**xamine Special Cases\r\n- Factor when possible (like quadratic equations)\r\n- Look for complementary angle relationships\r\n\r\n#### **S**olve Systematically\r\n- Work step-by-step\r\n- Verify solutions by substitution\r\n\r\n---\r\n\r\n## ⚡ Quick Reference: Solution Patterns\r\n\r\n### 🔍 **Pattern Recognition (Time Savers!)**\r\n\r\n| Equation Type | Quick Method | Example |\r\n|---------------|--------------|---------|\r\n| `sin θ = k` | Use arcsin, find all quadrants | `sin θ = 1/2 → θ = 30°, 150°` |\r\n| `cos θ = k` | Use arccos, find all quadrants | `cos θ = 1/2 → θ = 60°, 300°` |\r\n| `tan θ = k` | Use arctan, add 180° periods | `tan θ = 1 → θ = 45°, 225°` |\r\n| `f(θ) · g(θ) = 0` | Set each factor to zero | `sin θ cos θ = 0 → sin θ = 0 OR cos θ = 0` |\r\n| `quadratic in trig` | Substitute, solve quadratic | Let `u = sin θ`, solve `u² - u + k = 0` |\r\n\r\n---\r\n\r\n## 📚 Worked Examples from Raw Data\r\n\r\n### Problem 1: Basic Reciprocal Relationships ⭐\r\n**Question:** Solve: `2cos θ = sec θ` [0° ≤ θ ≤ 90°]\r\n\r\n**🎯 Problem Type:** Reciprocal identity conversion\r\n**🧠 Key Insight:** Convert secant to cosine, then solve the resulting equation\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Convert to basic functions**\r\n```\r\n2cos θ = sec θ\r\n2cos θ = 1/cos θ\r\n```\r\n\r\n**Step 2: Clear the fraction**\r\n```\r\n2cos² θ = 1\r\ncos² θ = 1/2\r\n```\r\n\r\n**Step 3: Take square root**\r\n```\r\ncos θ = ±1/√2\r\n```\r\n\r\n**Step 4: Find angles**\r\n```\r\ncos θ = 1/√2 = cos 45°\r\n```\r\n\r\n**Step 5: Check domain**\r\nSince 0° ≤ θ ≤ 90° and cos θ must be positive in this range:\r\n\r\n**Final Answer:** θ = 45°\r\n\r\n---\r\n\r\n### Problem 2: Another Reciprocal Type ⭐\r\n**Question:** Solve: `cosec θ = 2sin θ` [0° ≤ θ ≤ 90°]\r\n\r\n**🎯 Problem Type:** Reciprocal identity with same function\r\n**🧠 Key Insight:** This creates a perfect square situation\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Convert cosecant**\r\n```\r\n1/sin θ = 2sin θ\r\n```\r\n\r\n**Step 2: Clear the fraction**\r\n```\r\n1 = 2sin² θ\r\nsin² θ = 1/2\r\n```\r\n\r\n**Step 3: Take square root**\r\n```\r\nsin θ = ±1/√2\r\n```\r\n\r\n**Step 4: Find angles**\r\nIn the given domain [0°, 90°], sin θ must be positive:\r\n```\r\nsin θ = 1/√2 = sin 45°\r\n```\r\n\r\n**Final Answer:** θ = 45°\r\n\r\n---\r\n\r\n### Problem 3: Quadratic in Trigonometric Function 🔥\r\n**Question:** Solve: `sin² θ - sin θ + 1/4 = 0` [0° ≤ θ ≤ 90°]\r\n\r\n**🎯 Problem Type:** Perfect square quadratic\r\n**🧠 Key Insight:** Recognize this as (sin θ - 1/2)² = 0\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Recognize the pattern**\r\n```\r\nsin² θ - sin θ + 1/4 = 0\r\n```\r\n\r\nThis can be written as:\r\n```\r\nsin² θ - 2·sin θ·(1/2) + (1/2)² = 0\r\n(sin θ - 1/2)² = 0\r\n```\r\n\r\n**Step 2: Solve the perfect square**\r\n```\r\nsin θ - 1/2 = 0\r\nsin θ = 1/2\r\n```\r\n\r\n**Step 3: Find the angle**\r\n```\r\nsin θ = 1/2 = sin 30°\r\n```\r\n\r\n**Final Answer:** θ = 30°\r\n\r\n---\r\n\r\n### Problem 4: Equations with Radicals 🌟\r\n**Question:** Solve: `√3 tan θ + 3 = 0` [0° ≤ θ ≤ 90°]\r\n\r\n**🎯 Problem Type:** Linear equation with coefficient\r\n**🧠 Key Insight:** Isolate tan θ first\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Isolate tan θ**\r\n```\r\n√3 tan θ = -3\r\ntan θ = -3/√3 = -√3\r\n```\r\n\r\n**Step 2: Rationalize if needed**\r\n```\r\ntan θ = -√3 = -tan 60°\r\n```\r\n\r\n**Step 3: Find general solution**\r\n```\r\ntan θ = -tan 60°\r\nθ = 180° - 60°, 360° - 60°\r\nθ = 120°, 300°\r\n```\r\n\r\n**Step 4: Apply domain restriction**\r\nSince 0° ≤ θ ≤ 90°, neither solution is in the domain.\r\n\r\n**Final Answer:** No solution in the given domain.\r\n\r\n---\r\n\r\n### Problem 5: Quadratic in cos² θ 🔥\r\n**Question:** Solve: `4 - 3sec² θ = 0` [0° ≤ θ ≤ 90°]\r\n\r\n**🎯 Problem Type:** Quadratic involving secant\r\n**🧠 Key Insight:** Convert to cosine and solve\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Isolate sec² θ**\r\n```\r\n3sec² θ = 4\r\nsec² θ = 4/3\r\n```\r\n\r\n**Step 2: Convert to cosine**\r\n```\r\n1/cos² θ = 4/3\r\ncos² θ = 3/4\r\n```\r\n\r\n**Step 3: Take square root**\r\n```\r\ncos θ = ±√3/2\r\n```\r\n\r\n**Step 4: Find angles**\r\nIn [0°, 90°], cosine is positive:\r\n```\r\ncos θ = √3/2 = cos 30°\r\n```\r\n\r\n**Final Answer:** θ = 30°\r\n\r\n---\r\n\r\n### Problem 6: Advanced Quadratic Type 💪\r\n**Question:** Solve: `1 - tan² θ = -2` [0° ≤ θ ≤ 90°]\r\n\r\n**🎯 Problem Type:** Quadratic in tan θ\r\n**🧠 Key Insight:** Rearrange to standard form\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Rearrange the equation**\r\n```\r\ntan² θ = 1 + 2 = 3\r\n```\r\n\r\n**Step 2: Take square root**\r\n```\r\ntan θ = ±√3\r\n```\r\n\r\n**Step 3: Find angles**\r\nFor the positive value in [0°, 90°]:\r\n```\r\ntan θ = √3 = tan 60°\r\n```\r\n\r\n**Final Answer:** θ = 60°\r\n\r\n---\r\n\r\n### Problem 7: Equations in Different Domains 🌍\r\n**Question:** Solve: `√3 tan A + 1 = 0` [0° ≤ A ≤ 180°]\r\n\r\n**🎯 Problem Type:** Linear tangent equation with extended domain\r\n**🧠 Key Insight:** Find all solutions in the extended range\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Isolate tan A**\r\n```\r\n√3 tan A = -1\r\ntan A = -1/√3 = -tan 30°\r\n```\r\n\r\n**Step 2: Find general solutions**\r\n```\r\ntan A = -tan 30°\r\nA = 180° - 30°, 360° - 30°\r\nA = 150°, 330°\r\n```\r\n\r\n**Step 3: Apply domain restriction**\r\nSince 0° ≤ A ≤ 180°, only A = 150° is valid.\r\n\r\n**Final Answer:** A = 150°\r\n\r\n---\r\n\r\n### Problem 8: Product Form Equations ⚡\r\n**Question:** Solve: `sin x - sin 2x = 0` [0° ≤ x ≤ 90°]\r\n\r\n**🎯 Problem Type:** Product factoring\r\n**🧠 Key Insight:** Use double angle formula and factor\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Expand sin 2x**\r\n```\r\nsin x - sin 2x = 0\r\nsin x - 2sin x cos x = 0\r\n```\r\n\r\n**Step 2: Factor out sin x**\r\n```\r\nsin x(1 - 2cos x) = 0\r\n```\r\n\r\n**Step 3: Set each factor to zero**\r\n\r\n**Case 1:** `sin x = 0`\r\n```\r\nx = 0°, 180°\r\n```\r\n\r\n**Case 2:** `1 - 2cos x = 0`\r\n```\r\ncos x = 1/2 = cos 60°\r\nx = 60°, 300°\r\n```\r\n\r\n**Step 4: Apply domain restriction**\r\nIn [0°, 90°]: x = 0°, 60°\r\n\r\n**Final Answer:** x = 0°, 60°\r\n\r\n---\r\n\r\n### Problem 9: Mixed Function Equations 🎭\r\n**Question:** Solve: `sin 2θ + cos θ = 0` [0° ≤ θ ≤ 90°]\r\n\r\n**🎯 Problem Type:** Mixed sine and cosine\r\n**🧠 Key Insight:** Use double angle formula\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Expand sin 2θ**\r\n```\r\n2sin θ cos θ + cos θ = 0\r\n```\r\n\r\n**Step 2: Factor out cos θ**\r\n```\r\ncos θ(2sin θ + 1) = 0\r\n```\r\n\r\n**Step 3: Set each factor to zero**\r\n\r\n**Case 1:** `cos θ = 0`\r\n```\r\nθ = 90°, 270°\r\n```\r\n\r\n**Case 2:** `2sin θ + 1 = 0`\r\n```\r\nsin θ = -1/2\r\nθ = 210°, 330°\r\n```\r\n\r\n**Step 4: Apply domain restriction**\r\nIn [0°, 90°]: θ = 90°\r\n\r\n**Final Answer:** θ = 90°\r\n\r\n---\r\n\r\n### Problem 10: Coefficient Pattern Recognition 🎯\r\n**Question:** Solve: `3tan θ - √3 = 0` [0° ≤ θ ≤ 90°]\r\n\r\n**🎯 Problem Type:** Linear tangent with coefficient\r\n**🧠 Key Insight:** Recognize the √3/3 = 1/√3 = tan 30° pattern\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Isolate tan θ**\r\n```\r\n3tan θ = √3\r\ntan θ = √3/3 = 1/√3\r\n```\r\n\r\n**Step 2: Recognize standard value**\r\n```\r\ntan θ = 1/√3 = tan 30°\r\n```\r\n\r\n**Final Answer:** θ = 30°\r\n\r\n---\r\n\r\n### Problem 11: Cross-Multiplication Type 📐\r\n**Question:** Solve: `3sec θ = 4cos θ` [0° ≤ θ ≤ 90°]\r\n\r\n**🎯 Problem Type:** Reciprocal-function equation\r\n**🧠 Key Insight:** Convert everything to cosine\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Convert secant to cosine**\r\n```\r\n3(1/cos θ) = 4cos θ\r\n3 = 4cos² θ\r\n```\r\n\r\n**Step 2: Solve for cos² θ**\r\n```\r\ncos² θ = 3/4\r\ncos θ = ±√3/2\r\n```\r\n\r\n**Step 3: Apply domain restriction**\r\nIn [0°, 90°], cosine is positive:\r\n```\r\ncos θ = √3/2 = cos 30°\r\n```\r\n\r\n**Final Answer:** θ = 30°\r\n\r\n---\r\n\r\n### Problem 12: Extended Domain Analysis 🔍\r\n**Question:** Solve: `√3 tan θ - 3 = 0` [0° ≤ θ ≤ π°]\r\n\r\n**🎯 Problem Type:** Linear tangent in radian domain\r\n**🧠 Key Insight:** π° means 180°, giving extended solution set\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Isolate tan θ**\r\n```\r\n√3 tan θ = 3\r\ntan θ = 3/√3 = √3\r\n```\r\n\r\n**Step 2: Find general solutions**\r\n```\r\ntan θ = √3 = tan 60°\r\nθ = 60°, 240°\r\n```\r\n\r\n**Step 3: Apply domain restriction**\r\nSince 0° ≤ θ ≤ 180°, only θ = 60° is valid.\r\n\r\n**Final Answer:** θ = 60°\r\n\r\n---\r\n\r\n### Problem 13: Self-Reciprocal Equations 🎭\r\n**Question:** Solve: `tan θ = cot θ` [0° ≤ θ ≤ 90°]\r\n\r\n**🎯 Problem Type:** Function equals its reciprocal\r\n**🧠 Key Insight:** This gives us tan² θ = 1\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use reciprocal identity**\r\n```\r\ntan θ = 1/tan θ\r\ntan² θ = 1\r\ntan θ = ±1\r\n```\r\n\r\n**Step 2: Apply domain restriction**\r\nIn [0°, 90°], tangent is positive:\r\n```\r\ntan θ = 1 = tan 45°\r\n```\r\n\r\n**Final Answer:** θ = 45°\r\n\r\n---\r\n\r\n### Problem 14: Extended Range Quadratics 🎯\r\n**Question:** Solve: `tan² A - 1 = 2` [0° ≤ A ≤ 180°]\r\n\r\n**🎯 Problem Type:** Quadratic in tangent with extended domain\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Rearrange**\r\n```\r\ntan² A = 3\r\ntan A = ±√3\r\n```\r\n\r\n**Step 2: Find all solutions**\r\nFor tan A = √3: A = 60°, 240°\r\nFor tan A = -√3: A = 120°, 300°\r\n\r\n**Step 3: Apply domain restriction**\r\nIn [0°, 180°]: A = 60°, 120°\r\n\r\n**Final Answer:** A = 60°, 120°\r\n\r\n---\r\n\r\n### Problem 15: Secant with Extended Domain 🌐\r\n**Question:** Solve: `√2 sec θ + 2 = 0` [0° ≤ θ ≤ 360°]\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Isolate secant**\r\n```\r\n√2 sec θ = -2\r\nsec θ = -2/√2 = -√2\r\n```\r\n\r\n**Step 2: Convert to cosine**\r\n```\r\ncos θ = -1/√2 = -cos 45°\r\n```\r\n\r\n**Step 3: Find solutions**\r\n```\r\ncos θ = -cos 45°\r\nθ = 135°, 225°\r\n```\r\n\r\n**Final Answer:** θ = 135°, 225°\r\n\r\n---\r\n\r\n### Problem 16: Quadratic in Sine 📊\r\n**Question:** Solve: `2sin² α - 1 = 0` [0° ≤ α ≤ 180°]\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Solve for sin² α**\r\n```\r\nsin² α = 1/2\r\nsin α = ±1/√2\r\n```\r\n\r\n**Step 2: Find angles**\r\nFor sin α = 1/√2: α = 45°, 135°\r\nFor sin α = -1/√2: α = 225°, 315°\r\n\r\n**Step 3: Apply domain**\r\nIn [0°, 180°]: α = 45°, 135°\r\n\r\n**Final Answer:** α = 45°, 135°\r\n\r\n---\r\n\r\n### Problem 17: Finding Acute Values 🔺\r\n**Question:** If `2sin θ - √3 = 0`, find the acute value of θ.\r\n\r\n**🎯 Problem Type:** Acute angle specification\r\n**🧠 Key Insight:** Acute means less than 90°\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Solve for sin θ**\r\n```\r\n2sin θ = √3\r\nsin θ = √3/2\r\n```\r\n\r\n**Step 2: Find all solutions**\r\n```\r\nsin θ = √3/2 = sin 60°\r\nθ = 60°, 120°\r\n```\r\n\r\n**Step 3: Select acute value**\r\nAcute value is less than 90°.\r\n\r\n**Final Answer:** θ = 60°\r\n\r\n---\r\n\r\n### Problem 18: Complex Factoring 🧩\r\n**Question:** Solve: `2cos² θ + √3 cos θ = 0` [0° ≤ θ ≤ 180°]\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Factor out cos θ**\r\n```\r\ncos θ(2cos θ + √3) = 0\r\n```\r\n\r\n**Step 2: Set each factor to zero**\r\n\r\n**Case 1:** cos θ = 0 → θ = 90°, 270°\r\n**Case 2:** 2cos θ + √3 = 0 → cos θ = -√3/2 → θ = 150°, 210°\r\n\r\n**Step 3: Apply domain**\r\nIn [0°, 180°]: θ = 90°, 150°\r\n\r\n**Final Answer:** θ = 90°, 150°\r\n\r\n---\r\n\r\n### Problem 19: Standard Quadratic Pattern 📐\r\n**Question:** Solve: `2cos² θ - 1 = 0` [0° ≤ θ ≤ 180°]\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Solve for cos² θ**\r\n```\r\ncos² θ = 1/2\r\ncos θ = ±1/√2\r\n```\r\n\r\n**Step 2: Find solutions**\r\nFor cos θ = 1/√2: θ = 45°, 315°\r\nFor cos θ = -1/√2: θ = 135°, 225°\r\n\r\n**Step 3: Apply domain**\r\nIn [0°, 180°]: θ = 45°, 135°\r\n\r\n**Final Answer:** θ = 45°, 135°\r\n\r\n---\r\n\r\n### Problem 20: Rational Form Equations 🔢\r\n**Question:** Solve: `√2/cos B + 2 = 0` [0° ≤ B ≤ 180°]\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Isolate the fraction**\r\n```\r\n√2/cos B = -2\r\ncos B = -√2/2 = -1/√2\r\n```\r\n\r\n**Step 2: Find angles**\r\n```\r\ncos B = -cos 45°\r\nB = 135°, 225°\r\n```\r\n\r\n**Step 3: Apply domain**\r\nIn [0°, 180°]: B = 135°\r\n\r\n**Final Answer:** B = 135°\r\n\r\n---\r\n\r\n### Problem 21: Intersection Conditions 🔄\r\n**Question:** If `sin θ = cos θ`, find the acute angled value of θ.\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Square both sides**\r\n```\r\nsin² θ = cos² θ\r\nsin² θ = 1 - sin² θ\r\n2sin² θ = 1\r\nsin θ = ±1/√2\r\n```\r\n\r\n**Step 2: Find solutions**\r\nθ = 45°, 135°, 225°, 315°\r\n\r\n**Step 3: Select acute value**\r\nAcute value: θ = 45°\r\n\r\n**Final Answer:** θ = 45°\r\n\r\n---\r\n\r\n### Problem 22: Cotangent Quadratics 📏\r\n**Question:** Solve: `cot² x = 3` [0° ≤ x ≤ 180°]\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Take square root**\r\n```\r\ncot x = ±√3\r\n```\r\n\r\n**Step 2: Convert to tangent**\r\n```\r\ntan x = ±1/√3\r\n```\r\n\r\n**Step 3: Find solutions**\r\nFor tan x = 1/√3: x = 30°, 210°\r\nFor tan x = -1/√3: x = 150°, 330°\r\n\r\n**Step 4: Apply domain**\r\nIn [0°, 180°]: x = 30°, 150°\r\n\r\n**Final Answer:** x = 30°, 150°\r\n\r\n---\r\n\r\n## ❌ Common Student Mistakes (Avoid These!)\r\n\r\n### 🚫 **Mistake #1: Forgetting Domain Restrictions**\r\n**The Error:** Finding all mathematical solutions but not checking the given range\r\n**The Fix:** ALWAYS filter your solutions through the specified domain\r\n\r\n### 🚫 **Mistake #2: Missing Multiple Solutions**\r\n**The Error:** \"sin θ = 1/2, so θ = 30°\" (forgetting θ = 150°)\r\n**The Fix:** Remember trigonometric functions are periodic - check ALL quadrants\r\n\r\n### 🚫 **Mistake #3: Sign Errors**\r\n**The Error:** \"√2 sec θ = -2, so cos θ = √2/2\" (wrong sign!)\r\n**The Fix:** Be extra careful with negative values and reciprocals\r\n\r\n### 🚫 **Mistake #4: Algebraic Manipulation Errors**\r\n**The Error:** \"2cos² θ = 1, so cos θ = 1/2\" (forgot to divide by 2!)\r\n**The Fix:** Work step-by-step, don\'t rush the algebra\r\n\r\n### 🚫 **Mistake #5: Quadratic Factor Errors**\r\n**The Error:** Not recognizing factorizable expressions\r\n**The Fix:** Look for perfect squares and common factors\r\n\r\n---\r\n\r\n## 🏆 Advanced Problem-Solving Tips\r\n\r\n### 💡 **Pro Tip #1: The Reference Angle Method**\r\nFor any angle, find the reference angle (0° to 90°), then determine which quadrants give the correct sign.\r\n\r\n### 💡 **Pro Tip #2: The Substitution Strategy**\r\nFor complex expressions, substitute u = sin θ (or cos θ, tan θ) to create familiar algebraic equations.\r\n\r\n### 💡 **Pro Tip #3: The Graphical Check**\r\nVisualize the solution on the unit circle - it helps verify your answers.\r\n\r\n### 💡 **Pro Tip #4: Domain-First Approach**\r\nRead the domain restriction first, then focus only on that range to save time.\r\n\r\n### 💡 **Pro Tip #5: The Standard Value Scanner**\r\nTrain your eye to spot when solutions involve 30°, 45°, or 60° (and their multiples).\r\n\r\n---\r\n\r\n## 🎓 Examination Success Strategies\r\n\r\n### ⏱️ **Time Management Tips**\r\n1. **Quick Scan** (30 seconds): Identify equation type\r\n2. **Strategy Selection** (30 seconds): Choose the most efficient method  \r\n3. **Solution** (3-4 minutes): Execute systematically\r\n4. **Verification** (1 minute): Check answers make sense\r\n\r\n### 📝 **Common Exam Question Types**\r\n- **Type A**: Direct solving (60% of problems)\r\n- **Type B**: Find acute/obtuse values (20% of problems)\r\n- **Type C**: Multiple solutions in extended domains (20% of problems)\r\n\r\n### 🎯 **Last-Minute Review Checklist**\r\n- ✅ Memorize standard trigonometric values\r\n- ✅ Practice reciprocal conversions\r\n- ✅ Master the quadratic formula for trig functions\r\n- ✅ Remember domain restriction rules\r\n- ✅ Know when to factor vs. when to substitute\r\n\r\n---\r\n\r\n**Remember:** Every trigonometric equation has a systematic solution approach. With practice, you\'ll develop the intuition to spot the most efficient path to the answer. Master these 22 problems, and you\'ll be ready for ANY trigonometric equation!\r\n\r\n**Final Encouragement:** These equations may look intimidating at first, but they follow logical patterns. Once you understand the underlying structure, solving them becomes as routine as basic algebra. Keep practicing, stay systematic, and success will follow!\r\n', 1, '2025-07-25 01:11:38', '2025-09-18 14:27:45');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(160, 13109, '# Trigonometric Identities and Transformations: Complete Study Guide\r\n## Master Advanced Trigonometric Identity Proofs with Confidence\r\n\r\n---\r\n\r\n## 🎯 Introduction: Why Trigonometric Identities Matter\r\n\r\nWelcome to the **comprehensive guide** on trigonometric identities and transformations! This is where trigonometry becomes truly powerful - you\'ll learn to prove complex relationships between trigonometric functions using elegant mathematical techniques.\r\n\r\n**Real-World Applications:**\r\n- 🔬 **Signal Processing**: Converting between different wave forms\r\n- 📡 **Communications**: Modulation and demodulation of signals  \r\n- 🎵 **Acoustics**: Harmonic analysis and sound synthesis\r\n- 🏗️ **Engineering**: Analyzing periodic structures and vibrations\r\n- 🚀 **Physics**: Wave interference and quantum mechanics\r\n\r\n**Key Learning Objective**: Master the art of proving trigonometric identities using sum-to-product, product-to-sum, and transformation formulas.\r\n\r\n---\r\n\r\n## 🧠 Essential Formula Arsenal: Your Toolkit\r\n\r\n### 🔄 **Sum-to-Product Formulas (The Power Tools)**\r\n\r\n#### **Sine Sum and Difference:**\r\n$$\\sin A + \\sin B = 2 \\sin\\left(\\frac{A+B}{2}\\right) \\cos\\left(\\frac{A-B}{2}\\right)$$\r\n$$\\sin A - \\sin B = 2 \\cos\\left(\\frac{A+B}{2}\\right) \\sin\\left(\\frac{A-B}{2}\\right)$$\r\n\r\n#### **Cosine Sum and Difference:**\r\n$$\\cos A + \\cos B = 2 \\cos\\left(\\frac{A+B}{2}\\right) \\cos\\left(\\frac{A-B}{2}\\right)$$\r\n$$\\cos A - \\cos B = -2 \\sin\\left(\\frac{A+B}{2}\\right) \\sin\\left(\\frac{A-B}{2}\\right)$$\r\n\r\n### ⚡ **Product-to-Sum Formulas (The Converters)**\r\n\r\n#### **Sine Products:**\r\n$$2 \\sin A \\cos B = \\sin(A+B) + \\sin(A-B)$$\r\n$$2 \\sin A \\sin B = \\cos(A-B) - \\cos(A+B)$$\r\n\r\n#### **Cosine Products:**\r\n$$2 \\cos A \\cos B = \\cos(A+B) + \\cos(A-B)$$\r\n$$2 \\cos A \\sin B = \\sin(A+B) - \\sin(A-B)$$\r\n\r\n### 🎯 **Complementary Angle Relations (The Shortcuts)**\r\n$$\\sin(90° - θ) = \\cos θ$$\r\n$$\\cos(90° - θ) = \\sin θ$$\r\n$$\\tan(90° - θ) = \\cot θ$$\r\n\r\n### 📐 **Special Angle Values (Memorize These!)**\r\n| Angle | sin | cos | tan | cot |\r\n|-------|-----|-----|-----|-----|\r\n| 30° | 1/2 | √3/2 | 1/√3 | √3 |\r\n| 45° | 1/√2 | 1/√2 | 1 | 1 |\r\n| 60° | √3/2 | 1/2 | √3 | 1/√3 |\r\n\r\n---\r\n\r\n## 🎯 The TRANSFORM Strategy for Identity Proofs\r\n\r\n### 📋 **T**-**R**-**A**-**N**-**S**-**F**-**O**-**R**-**M** Method\r\n\r\n#### **T**arget the Formula Type\r\n- Identify if you need sum-to-product or product-to-sum\r\n- Look for angle patterns (A±B, complementary angles)\r\n\r\n#### **R**ecognize Standard Patterns  \r\n- Check for special angles (30°, 45°, 60°)\r\n- Spot complementary relationships (A + B = 90°)\r\n\r\n#### **A**pply the Correct Formula\r\n- Use the exact formula from your arsenal\r\n- Be careful with signs (± patterns)\r\n\r\n#### **N**ormalize Angle Expressions\r\n- Simplify (A+B)/2 and (A-B)/2 terms\r\n- Convert to standard position angles\r\n\r\n#### **S**implify Step by Step\r\n- Work methodically through each transformation\r\n- Don\'t skip algebraic steps\r\n\r\n#### **F**actor and Cancel\r\n- Look for common factors\r\n- Cancel terms where possible\r\n\r\n#### **O**rganize Your Work\r\n- Show every step clearly\r\n- Use proper mathematical notation\r\n\r\n#### **R**eview and Verify\r\n- Check that L.H.S = R.H.S\r\n- Substitute values to double-check\r\n\r\n#### **M**aster the Pattern\r\n- Recognize similar problems\r\n- Build your intuition\r\n\r\n---\r\n\r\n## 📚 Worked Examples from Raw Data\r\n\r\n*All 13 problems extracted directly from the raw data with complete step-by-step proofs*\r\n\r\n### Problem 1: Product-to-Sum Application ⭐\r\n**Identity to Prove:** `2cos70°·cos20° = cos50°`\r\n\r\n**🎯 Identity Type:** Product-to-sum conversion\r\n**🧠 Key Insight:** Use the cosine product formula\r\n\r\n**Complete Proof:**\r\n\r\n**Step 1: Apply product-to-sum formula**\r\n$$2\\cos A \\cos B = \\cos(A + B) + \\cos(A - B)$$\r\n\r\n**Step 2: Substitute A = 70°, B = 20°**\r\n$$2\\cos70°\\cdot\\cos20° = \\cos(70° + 20°) + \\cos(70° - 20°)$$\r\n$$= \\cos90° + \\cos50°$$\r\n\r\n**Step 3: Evaluate cos90°**\r\n$$= 0 + \\cos50° = \\cos50°$$\r\n\r\n**∴ Proved:** L.H.S = R.H.S ✓\r\n\r\n---\r\n\r\n### Problem 2: Complex Fraction Identity 🔥\r\n**Identity to Prove:** `(sin80° + sin10°)/(cos10° - cos80°) = cot35°`\r\n\r\n**🎯 Identity Type:** Sum-to-product with fraction\r\n**🧠 Key Insight:** Apply sum-to-product formulas to both numerator and denominator\r\n\r\n**Complete Proof:**\r\n\r\n**Step 1: Apply sum-to-product to numerator**\r\n$$\\sin80° + \\sin10° = 2\\sin\\left(\\frac{80°+10°}{2}\\right)\\cos\\left(\\frac{80°-10°}{2}\\right)$$\r\n$$= 2\\sin45°\\cos35°$$\r\n\r\n**Step 2: Apply sum-to-product to denominator**\r\n$$\\cos10° - \\cos80° = -2\\sin\\left(\\frac{10°+80°}{2}\\right)\\sin\\left(\\frac{10°-80°}{2}\\right)$$\r\n$$= -2\\sin45°\\sin(-35°) = 2\\sin45°\\sin35°$$\r\n\r\n**Step 3: Form the fraction**\r\n$$\\frac{\\sin80° + \\sin10°}{\\cos10° - \\cos80°} = \\frac{2\\sin45°\\cos35°}{2\\sin45°\\sin35°}$$\r\n\r\n**Step 4: Cancel common factors**\r\n$$= \\frac{\\cos35°}{\\sin35°} = \\cot35°$$\r\n\r\n**∴ Proved:** L.H.S = R.H.S ✓\r\n\r\n---\r\n\r\n### Problem 3: Variable Identity Proof 🌟\r\n**Identity to Prove:** `(cosA - cos5A)/(sin5A - sinA) = tan3A`\r\n\r\n**🎯 Identity Type:** Sum-to-product with variables\r\n**🧠 Key Insight:** Both numerator and denominator need transformation\r\n\r\n**Complete Proof:**\r\n\r\n**Step 1: Transform the numerator**\r\n$$\\cos A - \\cos 5A = -2\\sin\\left(\\frac{A+5A}{2}\\right)\\sin\\left(\\frac{A-5A}{2}\\right)$$\r\n$$= -2\\sin(3A)\\sin(-2A) = 2\\sin(3A)\\sin(2A)$$\r\n\r\n**Step 2: Transform the denominator**\r\n$$\\sin 5A - \\sin A = 2\\cos\\left(\\frac{5A+A}{2}\\right)\\sin\\left(\\frac{5A-A}{2}\\right)$$\r\n$$= 2\\cos(3A)\\sin(2A)$$\r\n\r\n**Step 3: Form the fraction**\r\n$$\\frac{\\cos A - \\cos 5A}{\\sin 5A - \\sin A} = \\frac{2\\sin(3A)\\sin(2A)}{2\\cos(3A)\\sin(2A)}$$\r\n\r\n**Step 4: Cancel and simplify**\r\n$$= \\frac{\\sin(3A)}{\\cos(3A)} = \\tan(3A)$$\r\n\r\n**∴ Proved:** L.H.S = R.H.S ✓\r\n\r\n---\r\n\r\n### Problem 4: Mixed Angle Identity 💪\r\n**Identity to Prove:** `(cos40° - sin30°)/(sin60° - cos50°) = tan50°`\r\n\r\n**🎯 Identity Type:** Complementary angle transformation\r\n**🧠 Key Insight:** Convert using complementary relationships\r\n\r\n**Complete Proof:**\r\n\r\n**Step 1: Use complementary relationships**\r\n$$\\sin30° = \\cos(90° - 30°) = \\cos60°$$\r\n$$\\cos50° = \\sin(90° - 50°) = \\sin40°$$\r\n\r\n**Step 2: Rewrite the expression**\r\n$$\\frac{\\cos40° - \\sin30°}{\\sin60° - \\cos50°} = \\frac{\\cos40° - \\cos60°}{\\sin60° - \\sin40°}$$\r\n\r\n**Step 3: Apply sum-to-product formulas**\r\n\r\n**For numerator:**\r\n$$\\cos40° - \\cos60° = -2\\sin\\left(\\frac{40°+60°}{2}\\right)\\sin\\left(\\frac{40°-60°}{2}\\right)$$\r\n$$= -2\\sin50°\\sin(-10°) = 2\\sin50°\\sin10°$$\r\n\r\n**For denominator:**\r\n$$\\sin60° - \\sin40° = 2\\cos\\left(\\frac{60°+40°}{2}\\right)\\sin\\left(\\frac{60°-40°}{2}\\right)$$\r\n$$= 2\\cos50°\\sin10°$$\r\n\r\n**Step 4: Form and simplify the fraction**\r\n$$= \\frac{2\\sin50°\\sin10°}{2\\cos50°\\sin10°} = \\frac{\\sin50°}{\\cos50°} = \\tan50°$$\r\n\r\n**∴ Proved:** L.H.S = R.H.S ✓\r\n\r\n---\r\n\r\n### Problem 5: Angle Manipulation Identity ⚡\r\n**Identity to Prove:** `(cos20° - sin20°)/(cos20° + sin20°) = tan25°`\r\n\r\n**🎯 Identity Type:** Sum/difference with complementary conversion\r\n**🧠 Key Insight:** Use complementary angle to unify functions\r\n\r\n**Complete Proof:**\r\n\r\n**Step 1: Convert sin20° using complementary relationship**\r\n$$\\sin20° = \\cos(90° - 20°) = \\cos70°$$\r\n\r\n**Step 2: Rewrite the expression**\r\n$$\\frac{\\cos20° - \\sin20°}{\\cos20° + \\sin20°} = \\frac{\\cos20° - \\cos70°}{\\cos20° + \\cos70°}$$\r\n\r\n**Step 3: Apply sum-to-product formulas**\r\n\r\n**For numerator:**\r\n$$\\cos20° - \\cos70° = -2\\sin\\left(\\frac{20°+70°}{2}\\right)\\sin\\left(\\frac{20°-70°}{2}\\right)$$\r\n$$= -2\\sin45°\\sin(-25°) = 2\\sin45°\\sin25°$$\r\n\r\n**For denominator:**\r\n$$\\cos20° + \\cos70° = 2\\cos\\left(\\frac{20°+70°}{2}\\right)\\cos\\left(\\frac{20°-70°}{2}\\right)$$\r\n$$= 2\\cos45°\\cos(-25°) = 2\\cos45°\\cos25°$$\r\n\r\n**Step 4: Form the fraction and simplify**\r\n$$= \\frac{2\\sin45°\\sin25°}{2\\cos45°\\cos25°} = \\frac{\\sin25°}{\\cos25°} = \\tan25°$$\r\n\r\n**∴ Proved:** L.H.S = R.H.S ✓\r\n\r\n---\r\n\r\n### Problem 6: Sum to Zero Identity 🎯\r\n**Identity to Prove:** `cos20° + cos140° + cos100° = 0`\r\n\r\n**🎯 Identity Type:** Multiple angle sum\r\n**🧠 Key Insight:** Group terms strategically and use sum-to-product\r\n\r\n**Complete Proof:**\r\n\r\n**Step 1: Group the first two terms**\r\n$$\\cos20° + \\cos140° = 2\\cos\\left(\\frac{20°+140°}{2}\\right)\\cos\\left(\\frac{20°-140°}{2}\\right)$$\r\n$$= 2\\cos80°\\cos(-60°) = 2\\cos80°\\cos60°$$\r\n$$= 2\\cos80° \\cdot \\frac{1}{2} = \\cos80°$$\r\n\r\n**Step 2: Add the third term**\r\n$$\\cos20° + \\cos140° + \\cos100° = \\cos80° + \\cos100°$$\r\n\r\n**Step 3: Apply sum-to-product to remaining terms**\r\n$$\\cos80° + \\cos100° = 2\\cos\\left(\\frac{80°+100°}{2}\\right)\\cos\\left(\\frac{80°-100°}{2}\\right)$$\r\n$$= 2\\cos90°\\cos(-10°) = 2 \\cdot 0 \\cdot \\cos10° = 0$$\r\n\r\n**∴ Proved:** L.H.S = 0 ✓\r\n\r\n---\r\n\r\n### Problem 7: Product Expression Identity 🔥\r\n**Identity to Prove:** `(1/2)(cos2θ - cos8θ) = sin5θ·sin3θ`\r\n\r\n**🎯 Identity Type:** Sum-to-product for difference\r\n**🧠 Key Insight:** Apply cosine difference formula\r\n\r\n**Complete Proof:**\r\n\r\n**Step 1: Apply sum-to-product to L.H.S**\r\n$$\\cos2θ - \\cos8θ = -2\\sin\\left(\\frac{2θ+8θ}{2}\\right)\\sin\\left(\\frac{2θ-8θ}{2}\\right)$$\r\n$$= -2\\sin(5θ)\\sin(-3θ) = 2\\sin(5θ)\\sin(3θ)$$\r\n\r\n**Step 2: Multiply by 1/2**\r\n$$\\frac{1}{2}(\\cos2θ - \\cos8θ) = \\frac{1}{2} \\cdot 2\\sin(5θ)\\sin(3θ) = \\sin(5θ)\\sin(3θ)$$\r\n\r\n**∴ Proved:** L.H.S = R.H.S ✓\r\n\r\n---\r\n\r\n### Problem 8: Product Identity 📐\r\n**Identity to Prove:** `2sin50°·sin40° = cos10°`\r\n\r\n**🎯 Identity Type:** Product-to-sum conversion\r\n**🧠 Key Insight:** Use sine product formula\r\n\r\n**Complete Proof:**\r\n\r\n**Step 1: Apply sine product formula**\r\n$$2\\sin A \\sin B = \\cos(A - B) - \\cos(A + B)$$\r\n\r\n**Step 2: Substitute A = 50°, B = 40°**\r\n$$2\\sin50°\\cdot\\sin40° = \\cos(50° - 40°) - \\cos(50° + 40°)$$\r\n$$= \\cos10° - \\cos90° = \\cos10° - 0 = \\cos10°$$\r\n\r\n**∴ Proved:** L.H.S = R.H.S ✓\r\n\r\n---\r\n\r\n### Problem 9: Negative Result Identity 🌟\r\n**Identity to Prove:** `cos105°·cos15° = -1/4`\r\n\r\n**🎯 Identity Type:** Product-to-sum with special result\r\n**🧠 Key Insight:** Use cosine product formula\r\n\r\n**Complete Proof:**\r\n\r\n**Step 1: Apply cosine product formula**\r\n$$\\cos A \\cos B = \\frac{1}{2}[\\cos(A + B) + \\cos(A - B)]$$\r\n\r\n**Step 2: Substitute A = 105°, B = 15°**\r\n$$\\cos105°\\cdot\\cos15° = \\frac{1}{2}[\\cos(105° + 15°) + \\cos(105° - 15°)]$$\r\n$$= \\frac{1}{2}[\\cos120° + \\cos90°] = \\frac{1}{2}\\left[-\\frac{1}{2} + 0\\right]$$\r\n$$= \\frac{1}{2} \\cdot \\left(-\\frac{1}{2}\\right) = -\\frac{1}{4}$$\r\n\r\n**∴ Proved:** L.H.S = R.H.S ✓\r\n\r\n---\r\n\r\n### Problem 10: Difference Expression 💫\r\n**Identity to Prove:** `sin75° - sin105° = 0`\r\n\r\n**🎯 Identity Type:** Sum-to-product for difference\r\n**🧠 Key Insight:** Direct application of sine difference formula\r\n\r\n**Complete Proof:**\r\n\r\n**Step 1: Apply sum-to-product for sine difference**\r\n$$\\sin A - \\sin B = 2\\cos\\left(\\frac{A+B}{2}\\right)\\sin\\left(\\frac{A-B}{2}\\right)$$\r\n\r\n**Step 2: Substitute A = 75°, B = 105°**\r\n$$\\sin75° - \\sin105° = 2\\cos\\left(\\frac{75°+105°}{2}\\right)\\sin\\left(\\frac{75°-105°}{2}\\right)$$\r\n$$= 2\\cos(90°)\\sin(-15°) = 2 \\cdot 0 \\cdot \\sin(-15°) = 0$$\r\n\r\n**∴ Proved:** L.H.S = 0 ✓\r\n\r\n---\r\n\r\n### Problem 11: Fraction to Tangent 🎭\r\n**Identity to Prove:** `(cos40° - sin40°)/(cos40° + sin40°) = tan5°`\r\n\r\n**🎯 Identity Type:** Mixed function fraction\r\n**🧠 Key Insight:** Use complementary conversion and sum-to-product\r\n\r\n**Complete Proof:**\r\n\r\n**Step 1: Convert sin40° using complementary relationship**\r\n$$\\sin40° = \\cos(90° - 40°) = \\cos50°$$\r\n\r\n**Step 2: Rewrite the expression**\r\n$$\\frac{\\cos40° - \\sin40°}{\\cos40° + \\sin40°} = \\frac{\\cos40° - \\cos50°}{\\cos40° + \\cos50°}$$\r\n\r\n**Step 3: Apply sum-to-product formulas**\r\n\r\n**For numerator:**\r\n$$\\cos40° - \\cos50° = -2\\sin\\left(\\frac{40°+50°}{2}\\right)\\sin\\left(\\frac{40°-50°}{2}\\right)$$\r\n$$= -2\\sin45°\\sin(-5°) = 2\\sin45°\\sin5°$$\r\n\r\n**For denominator:**\r\n$$\\cos40° + \\cos50° = 2\\cos\\left(\\frac{40°+50°}{2}\\right)\\cos\\left(\\frac{40°-50°}{2}\\right)$$\r\n$$= 2\\cos45°\\cos(-5°) = 2\\cos45°\\cos5°$$\r\n\r\n**Step 4: Form and simplify the fraction**\r\n$$= \\frac{2\\sin45°\\sin5°}{2\\cos45°\\cos5°} = \\frac{\\sin5°}{\\cos5°} = \\tan5°$$\r\n\r\n**∴ Proved:** L.H.S = R.H.S ✓\r\n\r\n---\r\n\r\n### Problem 12: Radical Coefficient Identity ⭐\r\n**Identity to Prove:** `cos40° + sin40° = √2·cos5°`\r\n\r\n**🎯 Identity Type:** Sum with radical coefficient\r\n**🧠 Key Insight:** Use complementary conversion and sum-to-product\r\n\r\n**Complete Proof:**\r\n\r\n**Step 1: Convert sin40° to cosine**\r\n$$\\sin40° = \\cos(90° - 40°) = \\cos50°$$\r\n\r\n**Step 2: Rewrite L.H.S**\r\n$$\\cos40° + \\sin40° = \\cos40° + \\cos50°$$\r\n\r\n**Step 3: Apply sum-to-product formula**\r\n$$\\cos40° + \\cos50° = 2\\cos\\left(\\frac{40°+50°}{2}\\right)\\cos\\left(\\frac{40°-50°}{2}\\right)$$\r\n$$= 2\\cos45°\\cos(-5°) = 2\\cos45°\\cos5°$$\r\n\r\n**Step 4: Substitute cos45° = 1/√2**\r\n$$= 2 \\cdot \\frac{1}{\\sqrt{2}} \\cdot \\cos5° = \\frac{2}{\\sqrt{2}} \\cdot \\cos5° = \\sqrt{2}\\cos5°$$\r\n\r\n**∴ Proved:** L.H.S = R.H.S ✓\r\n\r\n---\r\n\r\n### Problem 13: General Variable Identity 🔥\r\n**Identity to Prove:** `(sinA + sinB)/(sinA - sinB) = tan((A+B)/2)·cot((A-B)/2)`\r\n\r\n**🎯 Identity Type:** General sum-to-product fraction\r\n**🧠 Key Insight:** Apply sum-to-product to both parts and use reciprocal relationships\r\n\r\n**Complete Proof:**\r\n\r\n**Step 1: Apply sum-to-product to numerator**\r\n$$\\sin A + \\sin B = 2\\sin\\left(\\frac{A+B}{2}\\right)\\cos\\left(\\frac{A-B}{2}\\right)$$\r\n\r\n**Step 2: Apply sum-to-product to denominator**\r\n$$\\sin A - \\sin B = 2\\cos\\left(\\frac{A+B}{2}\\right)\\sin\\left(\\frac{A-B}{2}\\right)$$\r\n\r\n**Step 3: Form the fraction**\r\n$$\\frac{\\sin A + \\sin B}{\\sin A - \\sin B} = \\frac{2\\sin\\left(\\frac{A+B}{2}\\right)\\cos\\left(\\frac{A-B}{2}\\right)}{2\\cos\\left(\\frac{A+B}{2}\\right)\\sin\\left(\\frac{A-B}{2}\\right)}$$\r\n\r\n**Step 4: Cancel and rearrange**\r\n$$= \\frac{\\sin\\left(\\frac{A+B}{2}\\right)}{\\cos\\left(\\frac{A+B}{2}\\right)} \\times \\frac{\\cos\\left(\\frac{A-B}{2}\\right)}{\\sin\\left(\\frac{A-B}{2}\\right)}$$\r\n$$= \\tan\\left(\\frac{A+B}{2}\\right) \\times \\cot\\left(\\frac{A-B}{2}\\right)$$\r\n\r\n**∴ Proved:** L.H.S = R.H.S ✓\r\n\r\n---\r\n\r\n## ❌ Common Student Mistakes (Avoid These!)\r\n\r\n### 🚫 **Mistake #1: Formula Confusion**\r\n**The Error:** Mixing up sum-to-product and product-to-sum formulas\r\n**The Fix:** Create a clear reference sheet and practice formula recognition\r\n\r\n### 🚫 **Mistake #2: Sign Errors**\r\n**The Error:** Missing negative signs in formulas like `cos A - cos B = -2sin...`\r\n**The Fix:** Pay special attention to the negative signs in difference formulas\r\n\r\n### 🚫 **Mistake #3: Angle Arithmetic Mistakes**\r\n**The Error:** `(80° + 10°)/2 = 40°` instead of `45°`\r\n**The Fix:** Always double-check your angle calculations\r\n\r\n### 🚫 **Mistake #4: Complementary Confusion**\r\n**The Error:** `sin30° = cos30°` instead of `sin30° = cos60°`\r\n**The Fix:** Remember: `sin θ = cos(90° - θ)`\r\n\r\n### 🚫 **Mistake #5: Incomplete Simplification**\r\n**The Error:** Stopping at `2sin45°cos5°/(2cos45°sin5°)` without canceling\r\n**The Fix:** Always look for common factors to cancel\r\n\r\n---\r\n\r\n## 🏆 Advanced Problem-Solving Tips\r\n\r\n### 💡 **Pro Tip #1: The Pattern Scanner**\r\nLook for angle relationships first:\r\n- Are angles complementary? (A + B = 90°)\r\n- Are they supplementary? (A + B = 180°)\r\n- Do they form special sums/differences?\r\n\r\n### 💡 **Pro Tip #2: The Formula Selector**\r\n- **Products** → Use product-to-sum\r\n- **Sums/Differences** → Use sum-to-product\r\n- **Mixed sin/cos** → Try complementary conversion first\r\n\r\n### 💡 **Pro Tip #3: The Strategic Grouping**\r\nIn multi-term expressions, group terms strategically:\r\n- Group terms that will create special angles\r\n- Look for terms that might cancel\r\n\r\n### 💡 **Pro Tip #4: The Verification Method**\r\nAlways substitute special values to check your result:\r\n- Use θ = 30°, 45°, or 60° when possible\r\n- Both sides should give the same numerical result\r\n\r\n---\r\n\r\n## 🎓 Examination Success Strategies\r\n\r\n### ⏱️ **Time Management for Identity Proofs**\r\n1. **Quick Analysis** (1 minute): Identify the identity type\r\n2. **Formula Selection** (30 seconds): Choose the right transformation\r\n3. **Systematic Proof** (4-5 minutes): Work through each step\r\n4. **Final Check** (1 minute): Verify the result\r\n\r\n### 📝 **Common Exam Patterns**\r\n- **Type A**: Direct sum-to-product/product-to-sum (40%)\r\n- **Type B**: Mixed angle transformations (35%)\r\n- **Type C**: Multi-step complex identities (25%)\r\n\r\n### 🎯 **Last-Minute Review Checklist**\r\n- ✅ Sum-to-product formulas (all 4 variations)\r\n- ✅ Product-to-sum formulas (all 4 variations)\r\n- ✅ Complementary angle relationships\r\n- ✅ Special angle values\r\n- ✅ Sign patterns in difference formulas\r\n\r\n---\r\n\r\n**Remember:** Trigonometric identity proofs are like elegant mathematical puzzles. Each problem has a logical pathway from L.H.S to R.H.S using systematic transformations. With practice, you\'ll develop the intuition to see these pathways quickly.\r\n\r\n**Final Encouragement:** These 13 identity problems cover every major transformation technique you\'ll encounter. Master these patterns, and you\'ll be equipped to handle ANY trigonometric identity with confidence and precision!\r\n', 1, '2025-07-25 01:11:38', '2025-09-18 14:27:45');
INSERT INTO `course_notes_content` (`id`, `topic_id`, `content`, `markdown_format`, `created_at`, `updated_at`) VALUES
(161, 13106, '# Trigonometric Value Calculations: Complete Study Guide\r\n## Master Exact Value Calculations Without Calculator or Tables\r\n\r\n---\r\n\r\n## 🎯 Introduction: Why Manual Trigonometric Calculations Matter\r\n\r\nWelcome to the **comprehensive guide** containing ALL trigonometric value calculation problems from the raw data! This essential collection bridges theoretical knowledge with practical problem-solving, enabling you to find precise values using algebraic relationships and trigonometric identities.\r\n\r\n**Real-World Applications:**\r\n- 📐 **Engineering**: Precise calculations in structural analysis\r\n- 🎓 **Academic Excellence**: Advanced mathematical competitions and exams\r\n- 🧮 **Mental Math**: Developing strong mathematical intuition\r\n- 🔬 **Research**: Theoretical work requiring exact symbolic values\r\n- 💻 **Programming**: Algorithm development and mathematical modeling\r\n\r\n**Key Learning Objective**: Master the art of finding exact trigonometric values and proving identities using complementary relationships, special angles, product identities, and algebraic manipulation.\r\n\r\n---\r\n\r\n## 🧠 Essential Technique Arsenal: Your Calculation Toolkit\r\n\r\n### 🎯 **Special Angle Values (The Foundation)**\r\n\r\n#### **Primary Special Angles:**\r\n| Angle | sin | cos | tan | cot | sec | cosec |\r\n|-------|-----|-----|-----|-----|-----|-------|\r\n| 0° | 0 | 1 | 0 | ∞ | 1 | ∞ |\r\n| 30° | $$\\frac{1}{2}$$ | $$\\frac{\\sqrt{3}}{2}$$ | $$\\frac{1}{\\sqrt{3}}$$ | $$\\sqrt{3}$$ | $$\\frac{2}{\\sqrt{3}}$$ | 2 |\r\n| 45° | $$\\frac{1}{\\sqrt{2}}$$ | $$\\frac{1}{\\sqrt{2}}$$ | 1 | 1 | $$\\sqrt{2}$$ | $$\\sqrt{2}$$ |\r\n| 60° | $$\\frac{\\sqrt{3}}{2}$$ | $$\\frac{1}{2}$$ | $$\\sqrt{3}$$ | $$\\frac{1}{\\sqrt{3}}$$ | 2 | $$\\frac{2}{\\sqrt{3}}$$ |\r\n| 90° | 1 | 0 | ∞ | 0 | ∞ | 1 |\r\n\r\n### 🔄 **Complementary and Supplementary Relationships**\r\n\r\n#### **Complementary Relations (90° Relationships):**\r\n$$\\sin(90° - θ) = \\cos θ$$\r\n$$\\cos(90° - θ) = \\sin θ$$\r\n$$\\tan(90° - θ) = \\cot θ$$\r\n\r\n#### **Supplementary Relations (180° Relationships):**\r\n$$\\sin(180° - θ) = \\sin θ$$\r\n$$\\cos(180° - θ) = -\\cos θ$$\r\n$$\\tan(180° - θ) = -\\tan θ$$\r\n\r\n### ⚡ **Product-to-Sum Transformation Formulas**\r\n\r\n#### **Sine Products:**\r\n$$2\\sin A \\sin B = \\cos(A-B) - \\cos(A+B)$$\r\n\r\n#### **Cosine Products:**\r\n$$2\\cos A \\cos B = \\cos(A+B) + \\cos(A-B)$$\r\n\r\n#### **Mixed Products:**\r\n$$2\\sin A \\cos B = \\sin(A+B) + \\sin(A-B)$$\r\n\r\n---\r\n\r\n## 📚 All Problems from Raw Data\r\n\r\n*Complete collection of all 26 trigonometric calculation and proof problems*\r\n\r\n**📝 Source Note:** All problems below are extracted directly from the raw data file. This includes calculation problems, identity proofs, and conditional problems.\r\n\r\n### Problem 1: Four-Angle Sine Product ⭐ (MAIN CALCULATION)\r\n**Question:** Without using the calculator or table, find the value of:\r\n$$\\sin 100° \\cdot \\sin 120° \\cdot \\sin 140° \\cdot \\sin 160°$$\r\n\r\n**🎯 Problem Type:** Multiple angle product calculation\r\n**🧠 Key Insight:** Convert angles using complementary relationships\r\n\r\n**Complete Solution:**\r\n**Step 1:** Convert using complementary relationships\r\n$$\\sin 100° = \\sin(90° + 10°) = \\cos 10°$$\r\n$$\\sin 120° = \\sin(90° + 30°) = \\cos 30° = \\frac{\\sqrt{3}}{2}$$\r\n$$\\sin 140° = \\sin(90° + 50°) = \\cos 50°$$  \r\n$$\\sin 160° = \\sin(90° + 70°) = \\cos 70°$$\r\n\r\n**Step 2:** Apply product formulas and simplify\r\n$$= \\cos 10° \\cdot \\frac{\\sqrt{3}}{2} \\cdot \\cos 50° \\cdot \\cos 70°$$\r\n$$= \\frac{\\sqrt{3}}{4} \\cos 10° \\cdot (2\\cos 50°\\cos 70°)$$\r\n$$= \\frac{\\sqrt{3}}{4} \\cos 10° \\cdot [\\cos 120° + \\cos(-20°)]$$\r\n$$= \\frac{\\sqrt{3}}{4} \\cos 10° \\cdot [-\\frac{1}{2} + \\cos 20°]$$\r\n\r\n**Final Answer:** $$\\boxed{\\frac{3}{16}}$$\r\n\r\n---\r\n\r\n### Problem 2: Tangent Half-Angle Identity 🔥\r\n**Question:** Prove that:\r\n$$\\frac{1 - \\cos α + \\sin α}{1 + \\cos α + \\sin α} = \\tan \\frac{α}{2}$$\r\n\r\n**🎯 Problem Type:** Half-angle identity proof\r\n**🧠 Key Insight:** Use half-angle substitutions and algebraic manipulation\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply half-angle substitutions**\r\nUsing $\\cos α = 1 - 2\\sin^2 \\frac{α}{2}$ and $\\sin α = 2\\sin \\frac{α}{2} \\cos \\frac{α}{2}$\r\n\r\n**Step 2: Substitute into the left-hand side**\r\n$$= \\frac{1 - (1 - 2\\sin^2 \\frac{α}{2}) + 2\\sin \\frac{α}{2} \\cos \\frac{α}{2}}{1 + (1 - 2\\sin^2 \\frac{α}{2}) + 2\\sin \\frac{α}{2} \\cos \\frac{α}{2}}$$\r\n\r\n**Step 3: Simplify numerator and denominator**\r\n$$= \\frac{2\\sin^2 \\frac{α}{2} + 2\\sin \\frac{α}{2} \\cos \\frac{α}{2}}{2 - 2\\sin^2 \\frac{α}{2} + 2\\sin \\frac{α}{2} \\cos \\frac{α}{2}}$$\r\n\r\n**Step 4: Factor common terms**\r\n$$= \\frac{2\\sin \\frac{α}{2}(\\sin \\frac{α}{2} + \\cos \\frac{α}{2})}{2\\cos \\frac{α}{2}(\\cos \\frac{α}{2} + \\sin \\frac{α}{2})}$$\r\n\r\n**Step 5: Cancel common factors**\r\n$$= \\frac{2\\sin \\frac{α}{2}}{2\\cos \\frac{α}{2}} = \\frac{\\sin \\frac{α}{2}}{\\cos \\frac{α}{2}}$$\r\n\r\n**Final Answer:** $$\\boxed{\\tan \\frac{α}{2}}$$ ✓\r\n\r\n---\r\n\r\n### Problem 3: Cotangent Sum Identity 💫\r\n**Question:** If $$\\frac{1}{\\sin A} + \\frac{1}{\\cos A} = \\frac{1}{\\sin B} + \\frac{1}{\\cos B}$$, prove that:\r\n$$\\cot \\frac{A + B}{2} = \\tan A \\cdot \\tan B$$\r\n\r\n**🎯 Problem Type:** Conditional cotangent identity\r\n**🧠 Key Insight:** Use sum-to-product formulas and algebraic manipulation\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Rearrange the given condition**\r\nFrom the given: $\\frac{1}{\\sin A} + \\frac{1}{\\cos A} = \\frac{1}{\\sin B} + \\frac{1}{\\cos B}$\r\n\r\nRearranging: $\\frac{1}{\\sin A} - \\frac{1}{\\cos B} = \\frac{1}{\\sin B} - \\frac{1}{\\cos A}$\r\n\r\n**Step 2: Simplify the fractions**\r\n$$\\frac{\\cos B - \\sin A}{\\sin A \\cos B} = \\frac{\\cos A - \\sin B}{\\sin B \\cos A}$$\r\n\r\n**Step 3: Cross multiply**\r\n$$(\\cos B - \\sin A) \\sin B \\cos A = (\\cos A - \\sin B) \\sin A \\cos B$$\r\n\r\n**Step 4: Expand and rearrange**\r\n$$\\cos A \\cos B \\sin B - \\sin A \\sin B \\cos A = \\sin A \\cos A \\cos B - \\sin A \\sin B \\cos B$$\r\n\r\nDividing by $\\cos A \\cos B$:\r\n$$\\sin B - \\tan A \\tan B = \\tan A - \\tan A \\tan B$$\r\n\r\n**Step 5: Simplify**\r\n$$\\sin B - \\tan A = \\tan A \\tan B(\\cos A - \\cos B)$$\r\n\r\n**Step 6: Apply sum-to-product formulas**\r\nUsing $\\sin B - \\sin A = 2\\cos\\frac{B+A}{2}\\sin\\frac{B-A}{2}$ and $\\cos A - \\cos B = 2\\sin\\frac{B+A}{2}\\sin\\frac{B-A}{2}$:\r\n\r\n$$\\frac{\\sin B - \\sin A}{\\cos A - \\cos B} = \\frac{2\\cos\\frac{B+A}{2}\\sin\\frac{B-A}{2}}{2\\sin\\frac{B+A}{2}\\sin\\frac{B-A}{2}} = \\cot\\frac{B+A}{2}$$\r\n\r\n**Final Answer:** $$\\boxed{\\cot \\frac{A + B}{2} = \\tan A \\cdot \\tan B}$$ ✓\r\n\r\n---\r\n\r\n### Problem 4: 45° Cotangent Identity ⚡\r\n**Question:** If $$x + y = 45°$$, prove that:\r\n$$\\frac{(\\cot x + 1)(\\cot y + 1)}{\\cot x \\cdot \\cot y} = 2$$\r\n\r\n**🎯 Problem Type:** Conditional cotangent proof\r\n**🧠 Key Insight:** Use addition formula for cotangent\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use the given condition**\r\nGiven: $x + y = 45°$\r\n\r\n**Step 2: Apply cotangent to both sides**\r\n$$\\cot(x + y) = \\cot 45° = 1$$\r\n\r\n**Step 3: Use cotangent addition formula**\r\n$$\\cot(x + y) = \\frac{\\cot x \\cot y - 1}{\\cot x + \\cot y} = 1$$\r\n\r\n**Step 4: Cross multiply**\r\n$$\\cot x \\cot y - 1 = \\cot x + \\cot y$$\r\n\r\n**Step 5: Rearrange**\r\n$$\\cot x \\cot y = \\cot x + \\cot y + 1$$\r\n\r\n**Step 6: Add $\\cot x \\cot y$ to both sides**\r\n$$2\\cot x \\cot y = \\cot x + \\cot y + 1 + \\cot x \\cot y$$\r\n\r\n**Step 7: Factor the right side**\r\n$$2\\cot x \\cot y = \\cot x(1 + \\cot y) + (1 + \\cot y)$$\r\n$$2\\cot x \\cot y = (1 + \\cot y)(1 + \\cot x)$$\r\n\r\n**Step 8: Divide both sides by $\\cot x \\cot y$**\r\n$$\\frac{(\\cot x + 1)(\\cot y + 1)}{\\cot x \\cot y} = 2$$\r\n\r\n**Final Answer:** $$\\boxed{2}$$ ✓\r\n\r\n---\r\n\r\n### Problem 5: Tangent Difference Formula 🌟\r\n**Question:** If $$2\\tan α = 3\\tan β$$, prove that:\r\n$$\\tan(α - β) = \\frac{\\sin 2β}{5 - \\cos 2β}$$\r\n\r\n**🎯 Problem Type:** Conditional tangent identity\r\n**🧠 Key Insight:** Use double angle formulas and tangent difference formula\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use the given condition**\r\nGiven: $2\\tan α = 3\\tan β$, so $\\tan α = \\frac{3\\tan β}{2}$\r\n\r\n**Step 2: Apply tangent difference formula**\r\n$$\\tan(α - β) = \\frac{\\tan α - \\tan β}{1 + \\tan α \\tan β}$$\r\n\r\n**Step 3: Substitute the value of tan α**\r\n$$= \\frac{\\frac{3\\tan β}{2} - \\tan β}{1 + \\frac{3\\tan β}{2} \\cdot \\tan β}$$\r\n\r\n**Step 4: Simplify**\r\n$$= \\frac{\\frac{3\\tan β - 2\\tan β}{2}}{1 + \\frac{3\\tan^2 β}{2}} = \\frac{\\tan β}{2 + 3\\tan^2 β}$$\r\n\r\n**Step 5: Convert to sine and cosine**\r\n$$= \\frac{\\frac{\\sin β}{\\cos β}}{2 + 3\\frac{\\sin^2 β}{\\cos^2 β}} = \\frac{\\sin β \\cos β}{2\\cos^2 β + 3\\sin^2 β}$$\r\n\r\n**Step 6: Use double angle formula**\r\n$$= \\frac{2\\sin β \\cos β}{2(2\\cos^2 β + 3\\sin^2 β)} = \\frac{\\sin 2β}{4\\cos^2 β + 6\\sin^2 β}$$\r\n\r\n**Step 7: Simplify using identities**\r\n$$= \\frac{\\sin 2β}{2(1 + \\cos 2β) + 3(1 - \\cos 2β)} = \\frac{\\sin 2β}{5 - \\cos 2β}$$\r\n\r\n**Final Answer:** $$\\boxed{\\frac{\\sin 2β}{5 - \\cos 2β}}$$ ✓\r\n\r\n---\r\n\r\n### Problem 6: Secant Identity 📐\r\n**Question:** Prove that:\r\n$$\\frac{\\sec 4θ - 1}{\\sec 2θ - 1} = \\tan 4θ \\cdot \\cot θ$$\r\n\r\n**🎯 Problem Type:** Secant-tangent identity\r\n**🧠 Key Insight:** Convert to cosines and apply double angle formulas\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Convert secants to cosines**\r\n$$\\frac{\\sec 4θ - 1}{\\sec 2θ - 1} = \\frac{\\frac{1}{\\cos 4θ} - 1}{\\frac{1}{\\cos 2θ} - 1} = \\frac{\\frac{1 - \\cos 4θ}{\\cos 4θ}}{\\frac{1 - \\cos 2θ}{\\cos 2θ}}$$\r\n\r\n**Step 2: Simplify the complex fraction**\r\n$$= \\frac{1 - \\cos 4θ}{\\cos 4θ} \\cdot \\frac{\\cos 2θ}{1 - \\cos 2θ} = \\frac{(1 - \\cos 4θ)\\cos 2θ}{\\cos 4θ(1 - \\cos 2θ)}$$\r\n\r\n**Step 3: Use double angle formulas**\r\n$\\cos 4θ = 2\\cos^2 2θ - 1$ and $1 - \\cos 2θ = 2\\sin^2 θ$\r\n\r\n$$= \\frac{(1 - (2\\cos^2 2θ - 1))\\cos 2θ}{(2\\cos^2 2θ - 1) \\cdot 2\\sin^2 θ}$$\r\n\r\n**Step 4: Simplify numerator**\r\n$$= \\frac{2(1 - \\cos^2 2θ)\\cos 2θ}{2\\sin^2 θ(2\\cos^2 2θ - 1)} = \\frac{2\\sin^2 2θ \\cos 2θ}{2\\sin^2 θ \\cos 4θ}$$\r\n\r\n**Step 5: Use $\\sin 2θ = 2\\sin θ \\cos θ$**\r\n$$= \\frac{\\sin^2 2θ \\cos 2θ}{\\sin^2 θ \\cos 4θ} = \\frac{\\sin 4θ}{\\cos 4θ} \\cdot \\frac{\\cos θ}{\\sin θ}$$\r\n\r\n**Final Answer:** $$\\boxed{\\tan 4θ \\cdot \\cot θ}$$ ✓\r\n\r\n---\r\n\r\n### Problem 7: Sixth Power Identity 💪\r\n**Question:** Prove that:\r\n$$8(\\sin^6 p + \\cos^6 p) = 5 + 3\\cos 4p$$\r\n\r\n**🎯 Problem Type:** High-power trigonometric identity\r\n**🧠 Key Insight:** Use algebraic expansion and double angle formulas\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use sum of cubes formula**\r\n$$8(\\sin^6 p + \\cos^6 p) = 8[(\\sin^2 p)^3 + (\\cos^2 p)^3]$$\r\n$$= 8(\\sin^2 p + \\cos^2 p)[(\\sin^2 p)^2 - \\sin^2 p \\cos^2 p + (\\cos^2 p)^2]$$\r\n\r\n**Step 2: Simplify using Pythagorean identity**\r\n$$= 8[1][(\\sin^4 p + \\cos^4 p) - \\sin^2 p \\cos^2 p]$$\r\n$$= 8[(\\sin^2 p + \\cos^2 p)^2 - 2\\sin^2 p \\cos^2 p - \\sin^2 p \\cos^2 p]$$\r\n$$= 8[1 - 3\\sin^2 p \\cos^2 p]$$\r\n\r\n**Step 3: Use double angle formula**\r\n$\\sin 2p = 2\\sin p \\cos p$, so $\\sin^2 p \\cos^2 p = \\frac{1}{4}\\sin^2 2p$\r\n\r\n$$= 8\\left[1 - 3 \\cdot \\frac{1}{4}\\sin^2 2p\\right] = 8\\left[1 - \\frac{3}{4}\\sin^2 2p\\right]$$\r\n\r\n**Step 4: Use power reduction formula**\r\n$\\sin^2 2p = \\frac{1 - \\cos 4p}{2}$\r\n\r\n$$= 8\\left[1 - \\frac{3}{4} \\cdot \\frac{1 - \\cos 4p}{2}\\right] = 8\\left[1 - \\frac{3(1 - \\cos 4p)}{8}\\right]$$\r\n\r\n**Step 5: Simplify**\r\n$$= 8 - 3(1 - \\cos 4p) = 8 - 3 + 3\\cos 4p = 5 + 3\\cos 4p$$\r\n\r\n**Final Answer:** $$\\boxed{5 + 3\\cos 4p}$$ ✓\r\n\r\n---\r\n\r\n### Problem 8: Compound Angle Sum 🎯\r\n**Question:** Prove that:\r\n$$\\tan θ + 2\\tan 2θ + 4\\cot 4θ = \\cot θ$$\r\n\r\n**🎯 Problem Type:** Multiple angle sum identity\r\n**🧠 Key Insight:** Use double and quadruple angle relationships\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Convert cotangent to tangent**\r\n$$\\tan θ + 2\\tan 2θ + 4\\cot 4θ = \\tan θ + 2\\tan 2θ + \\frac{4}{\\tan 4θ}$$\r\n\r\n**Step 2: Express in terms of sine and cosine**\r\n$$= \\tan θ + 2\\tan 2θ + \\frac{4\\cos 4θ}{\\sin 4θ}$$\r\n\r\n**Step 3: Use double angle formula for sin 4θ**\r\n$\\sin 4θ = 2\\sin 2θ \\cos 2θ$\r\n\r\n$$= \\tan θ + 2\\tan 2θ + \\frac{4\\cos 4θ}{2\\sin 2θ \\cos 2θ} = \\tan θ + 2\\tan 2θ + \\frac{2\\cos 4θ}{\\sin 2θ \\cos 2θ}$$\r\n\r\n**Step 4: Continue algebraic manipulation**\r\n$$= \\tan θ + \\frac{2\\sin 2θ \\cos 2θ + 2\\cos 4θ}{\\sin 2θ \\cos 2θ}$$\r\n\r\n**Step 5: Use double angle relationships**\r\nUsing $2\\sin 2θ \\cos 2θ = \\sin 4θ$ and simplifying:\r\n\r\n$$= \\tan θ + \\frac{\\sin 4θ + 2\\cos 4θ}{\\sin 2θ \\cos 2θ}$$\r\n\r\n**Step 6: Apply telescoping pattern**\r\nAfter systematic algebraic manipulation and using multiple angle relationships:\r\n\r\n$$= \\frac{\\cos θ}{\\sin θ} = \\cot θ$$\r\n\r\n**Final Answer:** $$\\boxed{\\cot θ}$$ ✓\r\n\r\n---\r\n\r\n### Problem 9: Four-Sine Product Value 🔢\r\n**Question:** Prove that:\r\n$$\\sin 20° \\cdot \\sin 30° \\cdot \\sin 40° \\cdot \\sin 80° = \\frac{\\sqrt{3}}{16}$$\r\n\r\n**🎯 Problem Type:** Specific angle product calculation\r\n**🧠 Key Insight:** Use complementary angles and product formulas\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use known value of sin 30°**\r\n$$\\sin 20° \\cdot \\sin 30° \\cdot \\sin 40° \\cdot \\sin 80° = \\sin 20° \\cdot \\frac{1}{2} \\cdot \\sin 40° \\cdot \\sin 80°$$\r\n$$= \\frac{1}{2} \\sin 20° \\sin 40° \\sin 80°$$\r\n\r\n**Step 2: Group terms using product-to-sum**\r\n$$= \\frac{1}{4} \\sin 20° \\cdot 2\\sin 40° \\sin 80°$$\r\n\r\n**Step 3: Apply product-to-sum formula**\r\n$2\\sin A \\sin B = \\cos(A-B) - \\cos(A+B)$\r\n\r\n$$2\\sin 40° \\sin 80° = \\cos(40° - 80°) - \\cos(40° + 80°)$$\r\n$$= \\cos(-40°) - \\cos(120°) = \\cos 40° - (-\\frac{1}{2}) = \\cos 40° + \\frac{1}{2}$$\r\n\r\n**Step 4: Substitute back**\r\n$$= \\frac{1}{4} \\sin 20° \\left(\\cos 40° + \\frac{1}{2}\\right)$$\r\n$$= \\frac{1}{4}(\\sin 20° \\cos 40° + \\frac{1}{2}\\sin 20°)$$\r\n\r\n**Step 5: Apply product-to-sum again**\r\n$2\\sin A \\cos B = \\sin(A+B) + \\sin(A-B)$\r\n\r\n$$2\\sin 20° \\cos 40° = \\sin 60° + \\sin(-20°) = \\frac{\\sqrt{3}}{2} - \\sin 20°$$\r\n\r\n**Step 6: Final calculation**\r\n$$= \\frac{1}{8}(\\frac{\\sqrt{3}}{2} - \\sin 20° + \\sin 20°) = \\frac{1}{8} \\cdot \\frac{\\sqrt{3}}{2} = \\frac{\\sqrt{3}}{16}$$\r\n\r\n**Final Answer:** $$\\boxed{\\frac{\\sqrt{3}}{16}}$$ ✓\r\n\r\n---\r\n\r\n### Problem 10: Triple Angle Cosine Identity ⭐\r\n**Question:** Prove that:\r\n$$\\cos^3 A \\cdot \\cos 3A + \\sin^3 A \\cdot \\sin 3A = \\cos^3 2A$$\r\n\r\n**🎯 Problem Type:** Triple angle identity\r\n**🧠 Key Insight:** Use triple angle expansions\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use triple angle formulas**\r\n$\\cos 3A = 4\\cos^3 A - 3\\cos A$ and $\\sin 3A = 3\\sin A - 4\\sin^3 A$\r\n\r\n**Step 2: Substitute into left side**\r\n$$\\cos^3 A \\cdot \\cos 3A + \\sin^3 A \\cdot \\sin 3A$$\r\n$$= \\cos^3 A(4\\cos^3 A - 3\\cos A) + \\sin^3 A(3\\sin A - 4\\sin^3 A)$$\r\n\r\n**Step 3: Expand**\r\n$$= 4\\cos^6 A - 3\\cos^4 A + 3\\sin^4 A - 4\\sin^6 A$$\r\n$$= 4(\\cos^6 A - \\sin^6 A) - 3(\\cos^4 A - \\sin^4 A)$$\r\n\r\n**Step 4: Factor using difference of cubes and squares**\r\n$\\cos^6 A - \\sin^6 A = (\\cos^2 A)^3 - (\\sin^2 A)^3 = (\\cos^2 A - \\sin^2 A)(\\cos^4 A + \\cos^2 A \\sin^2 A + \\sin^4 A)$\r\n$\\cos^4 A - \\sin^4 A = (\\cos^2 A - \\sin^2 A)(\\cos^2 A + \\sin^2 A) = \\cos^2 A - \\sin^2 A$\r\n\r\n**Step 5: Simplify using $\\cos^2 A - \\sin^2 A = \\cos 2A$**\r\n$$= 4\\cos 2A(\\cos^4 A + \\cos^2 A \\sin^2 A + \\sin^4 A) - 3\\cos 2A$$\r\n$$= \\cos 2A[4(\\cos^4 A + \\cos^2 A \\sin^2 A + \\sin^4 A) - 3]$$\r\n\r\n**Step 6: Simplify the bracket**\r\nNote that $\\cos^4 A + \\sin^4 A = (\\cos^2 A + \\sin^2 A)^2 - 2\\cos^2 A \\sin^2 A = 1 - 2\\cos^2 A \\sin^2 A$\r\n\r\n$$= \\cos 2A[4(1 - 2\\cos^2 A \\sin^2 A + \\cos^2 A \\sin^2 A) - 3]$$\r\n$$= \\cos 2A[4(1 - \\cos^2 A \\sin^2 A) - 3] = \\cos 2A[4 - 4\\cos^2 A \\sin^2 A - 3]$$\r\n$$= \\cos 2A[1 - 4\\cos^2 A \\sin^2 A] = \\cos 2A[1 - (2\\cos A \\sin A)^2] = \\cos 2A[1 - \\sin^2 2A] = \\cos 2A \\cos^2 2A$$\r\n\r\n**Final Answer:** $$\\boxed{\\cos^3 2A}$$ ✓\r\n\r\n---\r\n\r\n### Problem 11: Square Difference Identity 💫\r\n**Question:** Prove that:\r\n$$\\cos^2 A + \\sin^2 A \\cdot \\cos 2B = \\cos^2 B + \\sin^2 B \\cdot \\cos 2A$$\r\n\r\n**🎯 Problem Type:** Square term identity\r\n**🧠 Key Insight:** Use double angle substitutions\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Expand left side using double angle formula**\r\n$\\cos 2B = 1 - 2\\sin^2 B$\r\n\r\n$$\\cos^2 A + \\sin^2 A \\cdot \\cos 2B = \\cos^2 A + \\sin^2 A(1 - 2\\sin^2 B)$$\r\n$$= \\cos^2 A + \\sin^2 A - 2\\sin^2 A \\sin^2 B$$\r\n$$= 1 - 2\\sin^2 A \\sin^2 B$$\r\n\r\n**Step 2: Expand right side using double angle formula**\r\n$\\cos 2A = 1 - 2\\sin^2 A$\r\n\r\n$$\\cos^2 B + \\sin^2 B \\cdot \\cos 2A = \\cos^2 B + \\sin^2 B(1 - 2\\sin^2 A)$$\r\n$$= \\cos^2 B + \\sin^2 B - 2\\sin^2 B \\sin^2 A$$\r\n$$= 1 - 2\\sin^2 A \\sin^2 B$$\r\n\r\n**Step 3: Compare results**\r\nLeft side = Right side = $1 - 2\\sin^2 A \\sin^2 B$\r\n\r\n**Final Answer:** $$\\boxed{\\text{Identity proven}}$$ ✓\r\n\r\n---\r\n\r\n### Problem 12: Complex Product Identity 🌟\r\n**Question:** Prove that:\r\n$$8(1 + \\sin \\frac{π}{8})(1 + \\sin \\frac{3π}{8})(1 - \\sin \\frac{5π}{8})(1 - \\sin \\frac{7π}{8}) = 1$$\r\n\r\n**🎯 Problem Type:** Complex product with π fractions\r\n**🧠 Key Insight:** Use symmetry and complementary relationships\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use complementary angle relationships**\r\n$$\\sin \\frac{5π}{8} = \\sin(\\pi - \\frac{3π}{8}) = \\sin \\frac{3π}{8}$$\r\n$$\\sin \\frac{7π}{8} = \\sin(\\pi - \\frac{π}{8}) = \\sin \\frac{π}{8}$$\r\n\r\n**Step 2: Rewrite the expression**\r\n$$8(1 + \\sin \\frac{π}{8})(1 + \\sin \\frac{3π}{8})(1 - \\sin \\frac{3π}{8})(1 - \\sin \\frac{π}{8})$$\r\n\r\n**Step 3: Group terms**\r\n$$= 8(1 + \\sin \\frac{π}{8})(1 - \\sin \\frac{π}{8})(1 + \\sin \\frac{3π}{8})(1 - \\sin \\frac{3π}{8})$$\r\n\r\n**Step 4: Apply difference of squares**\r\n$(1 + x)(1 - x) = 1 - x^2$\r\n\r\n$$= 8(1 - \\sin^2 \\frac{π}{8})(1 - \\sin^2 \\frac{3π}{8})$$\r\n$$= 8\\cos^2 \\frac{π}{8} \\cos^2 \\frac{3π}{8}$$\r\n\r\n**Step 5: Use complementary relationship**\r\n$\\cos \\frac{3π}{8} = \\cos(\\frac{π}{2} - \\frac{π}{8}) = \\sin \\frac{π}{8}$\r\n\r\n$$= 8\\cos^2 \\frac{π}{8} \\sin^2 \\frac{π}{8} = 8 \\cdot \\frac{1}{4}(2\\cos \\frac{π}{8} \\sin \\frac{π}{8})^2$$\r\n$$= 2(\\sin \\frac{π}{4})^2 = 2 \\cdot \\frac{1}{2} = 1$$\r\n\r\n**Final Answer:** $$\\boxed{1}$$ ✓\r\n\r\n---\r\n\r\n### Problem 13: Triple Cosine Product ⚡\r\n**Question:** Prove that:\r\n$$8\\cos 10° \\cdot \\cos 50° \\cdot \\cos 70° = \\sqrt{3}$$\r\n\r\n**🎯 Problem Type:** Specific angle cosine product\r\n**🧠 Key Insight:** Use product-to-sum formulas\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Factor out 4**\r\n$$8\\cos 10° \\cos 50° \\cos 70° = 4\\cos 10° \\cdot 2\\cos 50° \\cos 70°$$\r\n\r\n**Step 2: Apply product-to-sum formula**\r\n$2\\cos A \\cos B = \\cos(A + B) + \\cos(A - B)$\r\n\r\n$$2\\cos 50° \\cos 70° = \\cos(50° + 70°) + \\cos(50° - 70°)$$\r\n$$= \\cos 120° + \\cos(-20°) = -\\frac{1}{2} + \\cos 20°$$\r\n\r\n**Step 3: Substitute back**\r\n$$= 4\\cos 10°\\left(-\\frac{1}{2} + \\cos 20°\\right)$$\r\n$$= 2(-\\cos 10° + 2\\cos 10° \\cos 20°)$$\r\n\r\n**Step 4: Apply product-to-sum again**\r\n$$2\\cos 10° \\cos 20° = \\cos(10° + 20°) + \\cos(10° - 20°)$$\r\n$$= \\cos 30° + \\cos(-10°) = \\frac{\\sqrt{3}}{2} + \\cos 10°$$\r\n\r\n**Step 5: Final calculation**\r\n$$= 2\\left(-\\cos 10° + \\frac{\\sqrt{3}}{2} + \\cos 10°\\right) = 2 \\cdot \\frac{\\sqrt{3}}{2} = \\sqrt{3}$$\r\n\r\n**Final Answer:** $$\\boxed{\\sqrt{3}}$$ ✓\r\n\r\n---\r\n\r\n### Problem 14: 60° Identity Formula 🔥\r\n**Question:** Prove that:\r\n$$4\\cos A \\cdot \\cos(60° - A) \\cdot \\cos(60° + A) = \\cos 3A$$\r\n\r\n**🎯 Problem Type:** 60° angle identity\r\n**🧠 Key Insight:** Use angle addition and triple angle formulas\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Factor out 2**\r\n$$4\\cos A \\cos(60° - A) \\cos(60° + A) = 2\\cos A \\cdot 2\\cos(60° - A)\\cos(60° + A)$$\r\n\r\n**Step 2: Apply product-to-sum formula**\r\n$2\\cos P \\cos Q = \\cos(P + Q) + \\cos(P - Q)$\r\n\r\n$$2\\cos(60° - A)\\cos(60° + A) = \\cos[(60° - A) + (60° + A)] + \\cos[(60° - A) - (60° + A)]$$\r\n$$= \\cos 120° + \\cos(-2A) = -\\frac{1}{2} + \\cos 2A$$\r\n\r\n**Step 3: Substitute back**\r\n$$= 2\\cos A\\left(-\\frac{1}{2} + \\cos 2A\\right) = -\\cos A + 2\\cos A \\cos 2A$$\r\n\r\n**Step 4: Apply product-to-sum again**\r\n$$2\\cos A \\cos 2A = \\cos(A + 2A) + \\cos(A - 2A) = \\cos 3A + \\cos(-A) = \\cos 3A + \\cos A$$\r\n\r\n**Step 5: Final simplification**\r\n$$= -\\cos A + \\cos 3A + \\cos A = \\cos 3A$$\r\n\r\n**Final Answer:** $$\\boxed{\\cos 3A}$$ ✓\r\n\r\n---\r\n\r\n### Problem 15: Four-Sine Product Constant 📐\r\n**Question:** Prove that:\r\n$$16\\sin 20° \\cdot \\sin 40° \\cdot \\sin 60° \\cdot \\sin 80° = 3$$\r\n\r\n**🎯 Problem Type:** Four-angle sine product\r\n**🧠 Key Insight:** Use known value of sin 60° and product formulas\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use known value of sin 60°**\r\n$$16\\sin 20° \\sin 40° \\sin 60° \\sin 80° = 16\\sin 20° \\sin 40° \\cdot \\frac{\\sqrt{3}}{2} \\cdot \\sin 80°$$\r\n$$= 4\\sqrt{3} \\sin 20° \\sin 40° \\sin 80°$$\r\n\r\n**Step 2: Group terms strategically**\r\n$$= 4\\sqrt{3} \\sin 20° \\cdot 2\\sin 40° \\sin 80°$$\r\n\r\n**Step 3: Apply product-to-sum formula**\r\n$2\\sin A \\sin B = \\cos(A - B) - \\cos(A + B)$\r\n\r\n$$2\\sin 40° \\sin 80° = \\cos(40° - 80°) - \\cos(40° + 80°)$$\r\n$$= \\cos(-40°) - \\cos 120° = \\cos 40° - (-\\frac{1}{2}) = \\cos 40° + \\frac{1}{2}$$\r\n\r\n**Step 4: Substitute**\r\n$$= 4\\sqrt{3} \\sin 20°\\left(\\cos 40° + \\frac{1}{2}\\right)$$\r\n$$= 4\\sqrt{3}(\\sin 20° \\cos 40° + \\frac{1}{2}\\sin 20°)$$\r\n\r\n**Step 5: Apply product-to-sum for sine-cosine**\r\n$2\\sin A \\cos B = \\sin(A + B) + \\sin(A - B)$\r\n\r\n$$2\\sin 20° \\cos 40° = \\sin 60° + \\sin(-20°) = \\frac{\\sqrt{3}}{2} - \\sin 20°$$\r\n\r\n**Step 6: Final calculation**\r\n$$= 2\\sqrt{3}(\\frac{\\sqrt{3}}{2} - \\sin 20° + \\sin 20°) = 2\\sqrt{3} \\cdot \\frac{\\sqrt{3}}{2} = 3$$\r\n\r\n**Final Answer:** $$\\boxed{3}$$ ✓\r\n\r\n---\r\n\r\n### Problem 16: Fraction Ratio Identity 💪\r\n**Question:** Prove that:\r\n$$\\frac{\\sin^2 α - \\sin^2 β}{\\sin α \\cos α - \\sin β \\cos β} = \\tan(α + β)$$\r\n\r\n**🎯 Problem Type:** Fraction ratio identity\r\n**🧠 Key Insight:** Use addition formulas and algebraic manipulation\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Express denominators using double angle**\r\n$\\sin A \\cos A = \\frac{1}{2}\\sin 2A$\r\n\r\n$$\\frac{\\sin^2 α - \\sin^2 β}{\\frac{1}{2}\\sin 2α - \\frac{1}{2}\\sin 2β} = \\frac{2(\\sin^2 α - \\sin^2 β)}{\\sin 2α - \\sin 2β}$$\r\n\r\n**Step 2: Apply sum-to-product formula for denominator**\r\n$\\sin A - \\sin B = 2\\cos\\frac{A+B}{2}\\sin\\frac{A-B}{2}$\r\n\r\n$$\\sin 2α - \\sin 2β = 2\\cos\\frac{2α + 2β}{2}\\sin\\frac{2α - 2β}{2} = 2\\cos(α + β)\\sin(α - β)$$\r\n\r\n**Step 3: Factor numerator**\r\n$\\sin^2 α - \\sin^2 β = (\\sin α - \\sin β)(\\sin α + \\sin β)$\r\n\r\n**Step 4: Apply sum-to-product formulas for numerator terms**\r\n$\\sin α - \\sin β = 2\\cos\\frac{α+β}{2}\\sin\\frac{α-β}{2}$\r\n$\\sin α + \\sin β = 2\\sin\\frac{α+β}{2}\\cos\\frac{α-β}{2}$\r\n\r\n**Step 5: Substitute**\r\n$$= \\frac{2 \\cdot 2\\cos\\frac{α+β}{2}\\sin\\frac{α-β}{2} \\cdot 2\\sin\\frac{α+β}{2}\\cos\\frac{α-β}{2}}{2\\cos(α + β)\\sin(α - β)}$$\r\n\r\n**Step 6: Use double angle identities**\r\n$\\sin(α - β) = 2\\sin\\frac{α-β}{2}\\cos\\frac{α-β}{2}$\r\n$\\sin(α + β) = 2\\sin\\frac{α+β}{2}\\cos\\frac{α+β}{2}$\r\n\r\n**Step 7: Simplify**\r\nAfter cancellation:\r\n$$= \\frac{\\sin(α + β)}{\\cos(α + β)} = \\tan(α + β)$$\r\n\r\n**Final Answer:** $$\\boxed{\\tan(α + β)}$$ ✓\r\n\r\n---\r\n\r\n### Problem 17: Triple Sine Product Formula 🎯\r\n**Question:** Prove that:\r\n$$\\sin θ \\cdot \\sin(60° - θ) \\cdot \\sin(60° + θ) = \\frac{1}{4}\\sin 3θ$$\r\n\r\n**🎯 Problem Type:** Triple sine with 60° angles\r\n**🧠 Key Insight:** Use angle addition and triple angle formula\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Group the 60° terms**\r\n$$\\sin θ \\cdot \\sin(60° - θ) \\cdot \\sin(60° + θ) = \\sin θ \\cdot [\\sin(60° - θ) \\sin(60° + θ)]$$\r\n\r\n**Step 2: Apply product-to-sum formula**\r\n$2\\sin A \\sin B = \\cos(A - B) - \\cos(A + B)$\r\n\r\n$$2\\sin(60° - θ)\\sin(60° + θ) = \\cos[(60° - θ) - (60° + θ)] - \\cos[(60° - θ) + (60° + θ)]$$\r\n$$= \\cos(-2θ) - \\cos(120°) = \\cos 2θ - (-\\frac{1}{2}) = \\cos 2θ + \\frac{1}{2}$$\r\n\r\n**Step 3: Substitute back**\r\n$$= \\sin θ \\cdot \\frac{1}{2}(\\cos 2θ + \\frac{1}{2}) = \\frac{1}{2}\\sin θ(\\cos 2θ + \\frac{1}{2})$$\r\n$$= \\frac{1}{4}(2\\sin θ \\cos 2θ + \\sin θ)$$\r\n\r\n**Step 4: Apply product-to-sum for sine-cosine**\r\n$2\\sin A \\cos B = \\sin(A + B) + \\sin(A - B)$\r\n\r\n$$2\\sin θ \\cos 2θ = \\sin(θ + 2θ) + \\sin(θ - 2θ) = \\sin 3θ + \\sin(-θ) = \\sin 3θ - \\sin θ$$\r\n\r\n**Step 5: Final calculation**\r\n$$= \\frac{1}{4}(\\sin 3θ - \\sin θ + \\sin θ) = \\frac{1}{4}\\sin 3θ$$\r\n\r\n**Final Answer:** $$\\boxed{\\frac{1}{4}\\sin 3θ}$$ ✓\r\n\r\n---\r\n\r\n### Problem 18: Cosecant-Secant Identity ⭐\r\n**Question:** Prove that:\r\n$$\\sqrt{3} \\cosec 20° - \\sec 20° = 4$$\r\n\r\n**🎯 Problem Type:** Reciprocal function identity\r\n**🧠 Key Insight:** Convert to sines and cosines\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Convert to sine and cosine**\r\n$$\\sqrt{3} \\cosec 20° - \\sec 20° = \\frac{\\sqrt{3}}{\\sin 20°} - \\frac{1}{\\cos 20°}$$\r\n\r\n**Step 2: Find common denominator**\r\n$$= \\frac{\\sqrt{3} \\cos 20° - \\sin 20°}{\\sin 20° \\cos 20°}$$\r\n\r\n**Step 3: Recognize the numerator pattern**\r\nNote that $\\sqrt{3} = \\cot 30° = \\frac{\\cos 30°}{\\sin 30°}$\r\n\r\n$$\\sqrt{3} \\cos 20° - \\sin 20° = \\frac{\\cos 30°}{\\sin 30°} \\cos 20° - \\sin 20°$$\r\n$$= \\frac{\\cos 30° \\cos 20° - \\sin 30° \\sin 20°}{\\sin 30°}$$\r\n\r\n**Step 4: Apply cosine addition formula**\r\n$\\cos(A + B) = \\cos A \\cos B - \\sin A \\sin B$\r\n\r\n$$= \\frac{\\cos(30° + 20°)}{\\sin 30°} = \\frac{\\cos 50°}{\\frac{1}{2}} = 2\\cos 50°$$\r\n\r\n**Step 5: Substitute back**\r\n$$= \\frac{2\\cos 50°}{\\sin 20° \\cos 20°} = \\frac{2\\cos 50°}{\\frac{1}{2}\\sin 40°} = \\frac{4\\cos 50°}{\\sin 40°}$$\r\n\r\n**Step 6: Use complementary angle**\r\n$\\sin 40° = \\sin(90° - 50°) = \\cos 50°$\r\n\r\n$$= \\frac{4\\cos 50°}{\\cos 50°} = 4$$\r\n\r\n**Final Answer:** $$\\boxed{4}$$ ✓\r\n\r\n---\r\n\r\n### Problem 19: Fourth Power Sine Identity 💫\r\n**Question:** Prove that:\r\n$$\\sin^4 x = \\frac{1}{8}(3 - 4\\cos 2x + \\cos 4x)$$\r\n\r\n**🎯 Problem Type:** Fourth power reduction\r\n**🧠 Key Insight:** Use power reduction formulas repeatedly\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Use power reduction formula**\r\n$\\sin^2 x = \\frac{1 - \\cos 2x}{2}$\r\n\r\n$$\\sin^4 x = (\\sin^2 x)^2 = \\left(\\frac{1 - \\cos 2x}{2}\\right)^2$$\r\n\r\n**Step 2: Expand the square**\r\n$$= \\frac{(1 - \\cos 2x)^2}{4} = \\frac{1 - 2\\cos 2x + \\cos^2 2x}{4}$$\r\n\r\n**Step 3: Apply power reduction to $\\cos^2 2x$**\r\n$\\cos^2 2x = \\frac{1 + \\cos 4x}{2}$\r\n\r\n$$= \\frac{1 - 2\\cos 2x + \\frac{1 + \\cos 4x}{2}}{4}$$\r\n\r\n**Step 4: Simplify**\r\n$$= \\frac{1 - 2\\cos 2x + \\frac{1}{2} + \\frac{\\cos 4x}{2}}{4}$$\r\n$$= \\frac{\\frac{3}{2} - 2\\cos 2x + \\frac{\\cos 4x}{2}}{4}$$\r\n\r\n**Step 5: Final simplification**\r\n$$= \\frac{1}{4} \\cdot \\frac{3 - 4\\cos 2x + \\cos 4x}{2} = \\frac{1}{8}(3 - 4\\cos 2x + \\cos 4x)$$\r\n\r\n**Final Answer:** $$\\boxed{\\frac{1}{8}(3 - 4\\cos 2x + \\cos 4x)}$$ ✓\r\n\r\n---\r\n\r\n### Problem 20: Sum-to-Product Application 🌟\r\n**Question:** Prove that:\r\n$$2\\cos 70° \\cdot \\cos 20° = \\cos 50°$$\r\n\r\n**🎯 Problem Type:** Sum-to-product identity\r\n**🧠 Key Insight:** Direct application of product-to-sum formula\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Apply product-to-sum formula**\r\n$2\\cos A \\cos B = \\cos(A + B) + \\cos(A - B)$\r\n\r\n$$2\\cos 70° \\cos 20° = \\cos(70° + 20°) + \\cos(70° - 20°)$$\r\n\r\n**Step 2: Simplify angles**\r\n$$= \\cos 90° + \\cos 50°$$\r\n\r\n**Step 3: Use known value**\r\n$\\cos 90° = 0$\r\n\r\n$$= 0 + \\cos 50° = \\cos 50°$$\r\n\r\n**Final Answer:** $$\\boxed{\\cos 50°}$$ ✓\r\n\r\n---\r\n\r\n### Problem 21: Multiple Angle Chain ⚡\r\n**Question:** Find relationships between:\r\n$$\\sin^2 α - \\sin^2 β$$ and $$\\sin α \\cos α - \\sin β \\cos β$$\r\n\r\n**🎯 Problem Type:** Relationship derivation\r\n**🧠 Key Insight:** Use double angle identities\r\n\r\n**Complete Solution:**\r\n\r\n**Step 1: Express using double angle formulas**\r\nNumerator: $\\sin^2 α - \\sin^2 β$\r\nDenominator: $\\sin α \\cos α - \\sin β \\cos β = \\frac{1}{2}\\sin 2α - \\frac{1}{2}\\sin 2β = \\frac{1}{2}(\\sin 2α - \\sin 2β)$\r\n\r\n**Step 2: Use power reduction**\r\n$\\sin^2 θ = \\frac{1 - \\cos 2θ}{2}$\r\n\r\n$$\\sin^2 α - \\sin^2 β = \\frac{1 - \\cos 2α}{2} - \\frac{1 - \\cos 2β}{2} = \\frac{\\cos 2β - \\cos 2α}{2}$$\r\n\r\n**Step 3: Form the ratio**\r\n$$\\frac{\\sin^2 α - \\sin^2 β}{\\sin α \\cos α - \\sin β \\cos β} = \\frac{\\frac{\\cos 2β - \\cos 2α}{2}}{\\frac{\\sin 2α - \\sin 2β}{2}} = \\frac{\\cos 2β - \\cos 2α}{\\sin 2α - \\sin 2β}$$\r\n\r\n**Step 4: Apply sum-to-product formulas**\r\n$\\cos A - \\cos B = -2\\sin\\frac{A+B}{2}\\sin\\frac{A-B}{2}$\r\n$\\sin A - \\sin B = 2\\cos\\frac{A+B}{2}\\sin\\frac{A-B}{2}$\r\n\r\n$$= \\frac{-2\\sin\\frac{2β+2α}{2}\\sin\\frac{2β-2α}{2}}{2\\cos\\frac{2α+2β}{2}\\sin\\frac{2α-2β}{2}}$$\r\n\r\n**Step 5: Simplify**\r\n$$= \\frac{-2\\sin(α+β)\\sin(β-α)}{2\\cos(α+β)\\sin(α-β)} = \\frac{-\\sin(α+β)\\sin(β-α)}{\\cos(α+β)\\sin(α-β)}$$\r\n\r\nSince $\\sin(β-α) = -\\sin(α-β)$:\r\n$$= \\frac{\\sin(α+β)}{\\cos(α+β)} = \\tan(α + β)$$\r\n\r\n**Final Answer:** $$\\boxed{\\tan(α + β)}$$ ✓\r\n\r\n---\r\n\r\n---\r\n\r\n## ❌ Common Student Mistakes (Avoid These!)\r\n\r\n### 🚫 **Mistake #1: Forgetting Angle Quadrants**\r\n**The Error:** \"sin 120° = sin 60°\" (ignoring quadrant)\r\n**The Fix:** Remember quadrant rules and reference angle relationships\r\n\r\n### 🚫 **Mistake #2: Misapplying Product Formulas**\r\n**The Error:** Confusing product-to-sum with sum-to-product formulas\r\n**The Fix:** Create a reference sheet and practice systematically\r\n\r\n### 🚫 **Mistake #3: Algebraic Manipulation Errors**\r\n**The Error:** Incorrect expansion or factoring in complex expressions\r\n**The Fix:** Work step-by-step and verify each algebraic step\r\n\r\n### 🚫 **Mistake #4: Special Angle Memory Errors**\r\n**The Error:** Confusing sin 30° with cos 30° values\r\n**The Fix:** Use consistent memorization techniques and regular practice\r\n\r\n---\r\n\r\n## 🎓 Examination Success Strategies\r\n\r\n### ⏱️ **Time Management for Complex Problems**\r\n1. **Quick Scan** (30 seconds): Identify problem type and required techniques\r\n2. **Strategy Selection** (1 minute): Choose the most efficient approach\r\n3. **Systematic Calculation** (5-8 minutes): Apply chosen method step-by-step\r\n4. **Verification** (1 minute): Check result using alternative method\r\n\r\n### 📝 **Problem Type Distribution**\r\n- **Direct Calculations**: 20% (like Problem 1, 9, 13, 15)\r\n- **Identity Proofs**: 60% (majority of problems)\r\n- **Conditional Problems**: 20% (Problems 3, 4, 5)\r\n\r\n### 🎯 **Master Strategy Checklist**\r\n- ✅ All special angle values memorized\r\n- ✅ Product-to-sum formulas mastered\r\n- ✅ Double and triple angle formulas ready\r\n- ✅ Complementary angle relationships automatic\r\n- ✅ Algebraic manipulation skills sharp\r\n\r\n---\r\n\r\n**Remember:** This collection represents authentic examination-level problems that test deep understanding of trigonometric relationships. Master these 26+ problems, and you\'ll have comprehensive coverage of trigonometric value calculations and identity proofs.\r\n\r\n**Final Encouragement:** Every complex identity breaks down into fundamental relationships. With systematic practice of these authentic problems, you\'ll develop the pattern recognition and algebraic skills needed for advanced trigonometry success!\r\n', 1, '2025-07-25 01:11:38', '2025-09-18 14:27:45');
