WAEC Further Mathematics 2018 Paper

Question : 18

Total: 40

\sin\theta = \frac{3}{5}, 0^{\circ} < \theta < 90^{\circ}

\cos(180 - \theta)

Solution:

Concept:

Use the Pythagorean triple and the cosine rule for supplementary angles:

\cos(180^\circ - \theta) = -\cos\theta

Explanation:

Given

\sin\theta = \frac{3}{5}

and

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

\theta

is in the first quadrant.

In a right triangle,

\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}

, so opposite side

= 3

and hypotenuse

= 5

Using Pythagoras: adjacent side

= \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4

Thus,

\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5}

Now,

180^\circ - \theta

lies in the second quadrant, where cosine is negative.

Since

\cos(180^\circ - \theta) = -\cos\theta = -\frac{4}{5}

Answer:

\cos(180^\circ - \theta) = -\frac{4}{5}

, so the correct option is D. $\frac{-4}{5}$ .

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