WAEC Further Mathematics 2018 Paper

$F_{1} = 7i + 0j$

$F_{2} = (4\cos\theta)i + (4\sin\theta)j$

$9 = \sqrt{(7 + 4\cos\theta)^{2} + (4\sin\theta)^{2}}$

$9^{2} = (7 + 4\cos\theta)^{2} + (4\sin\theta)^{2} \implies 81 = 49 + 56\cos\theta + 16\cos^{2}\theta + 16\sin^{2}\theta$

$81 = 49 + 56\cos\theta + 16(\cos^{2}\theta + \sin^{2}\theta)$

Recall, $\cos^{2}\theta + \sin^{2}\theta = 1$

$81 = 49 + 56\cos\theta + 16 \implies 81 - 49 - 16 = 56\cos\theta$

$16 = 56\cos\theta \implies \cos\theta = \frac{16}{56} = 0.2857$

$\theta = \cos^{-1} 0.2857 = 73.40^{\circ}$