WAEC 2023 · Paper 1 · Q25

Solve $6\sin 2\theta \tan\theta = 4$ , where $0^\circ < \theta < 90^\circ$ .

Worked solution (try it first)

Use $\sin 2\theta = 2\sin\theta\cos\theta$ and $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ : the left side is $12\sin\theta\cos\theta \cdot \dfrac{\sin\theta}{\cos\theta} = 12\sin^2\theta$ .
So $12\sin^2\theta = 4$ , which gives $\sin^2\theta = \frac{1}{3}$ .
$\theta$ is acute, so $\sin\theta = \frac{1}{\sqrt{3}} = 0.5774$ .
So $\theta = 35.26^\circ$ , option A.