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WAEC Mathematics 2016 Paper

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Question : 19

Total: 49

If cos

\theta

= x and sin 60^o = x + 0.5 0^o <

\theta

< 90^o, find, correct to the nearest degree, the value of

\theta

32^o
40^o
60^o
69^o

Solution:

Concept:

Use the standard value of

\sin 60^\circ

to first find

x

, then apply inverse cosine to determine

\theta

.

Explanation:

Given that

\sin 60^\circ = x + 0.5

.

We know

\sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.8660

.

So,

x + 0.5 = 0.8660

.

Subtract

0.5

from both sides:

x = 0.8660 - 0.5 = 0.3660

.

It is also given that

\cos \theta = x

.

Thus,

\cos \theta = 0.3660

.

Since

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

,

\theta

is acute.

Therefore,

\theta = \cos^{-1}(0.3660)

.

This gives

\theta \approx 68.53^\circ

.

Correct to the nearest degree,

\theta = 69^\circ

.

Answer:

\theta = 69^\circ

, which is option D.

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