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WAEC Further Mathematics 2023 Paper

Question : 19

Total: 40

Solve 6 sin 2θ tan θ = 4, where 0º < θ < 90º

Solution:

Concept:

Use the double-angle identity and the tangent identity to simplify the equation to a sine function.

Explanation:

Given:

6 \sin 2\theta \tan \theta = 4

, where

0^\circ < \theta < 90^\circ

Write

\sin 2\theta = 2\sin\theta\cos\theta

and

\tan\theta = \frac{\sin\theta}{\cos\theta}

Substitute these into the equation:

6 \times 2\sin\theta\cos\theta \times \frac{\sin\theta}{\cos\theta} = 4

The

\cos\theta

terms cancel, giving:

12\sin^2\theta = 4

Divide both sides by

12

\sin^2\theta = \frac{4}{12} = \frac{1}{3}

Take the positive square root since

\theta

is acute:

\sin\theta = \frac{1}{\sqrt{3}}

Therefore:

\theta = \sin^{-1}\left(\frac{1}{\sqrt{3}}\right) \approx 35.26^\circ

Answer:

\theta = 35.26^\circ

, which is option C.

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