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WAEC Further Mathematics 2012 Paper

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Question : 5

Total: 40

Given that

\sin \; x = \frac{-\sqrt{3}}{2}

\cos x > 0

, find x.

300°
240°
120°
60°

Solution:

Concept:

The quadrant is chosen using the signs of

\sin x

and

\cos x

.

Explanation:

Here,

\sin x = \frac{-\sqrt{3}}{2}

is negative, and

\cos x > 0

is positive.

This means

x

lies in the fourth quadrant, where

270^{\circ} < x < 360^{\circ}

.

The reference angle for

\frac{\sqrt{3}}{2}

is

60^{\circ}

.

In the fourth quadrant, the angle is

360^{\circ} - 60^{\circ} = 300^{\circ}

.

Check:

\sin 300^{\circ} = -\sin 60^{\circ} = \frac{-\sqrt{3}}{2}

, and

\cos 300^{\circ} = \cos 60^{\circ} = \frac{1}{2}

. Therefore,

x = 300^{\circ}

.

Answer:

300^{\circ}

(Option A)

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