If the resultant is perpendicular to F₂, then from the vector triangle, $\cos(90) = (F_1^2 + R^2 - F_2^2)/(2F_1R)$. Also, $F_1 + F_2 \cos\theta = 0$ (component of resultant along F1 is zero). With $R=F1$, this leads to $F_1 = -F_2 \cos\theta$. And $R^2 = F_1^2 + F_2^2 + 2F_1F_2\cos\theta$. This setup seems flawed. A common interpretation for such problems is that the resultant F is perpendicular to one of the vectors (say F2). Then $F_1\cos\alpha = -F_2$. Resultant along perpendicular direction is $R = F_1\sin\alpha$. If $R=F_1$, then $\sin\alpha=1$, so $\alpha=90$. Angle between F1 and F2 is 120.