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WAEC Mathematics 2021 Paper

Question : 38

Total: 49

Given that sin x = 3/5, 0 ≤ x ≤ 90, evaluate (tanx + 2cosx)

Solution:

Concept:

Use the right-triangle trigonometric ratios and the Pythagoras theorem.

Explanation:

Given

\sin x = \frac{3}{5} = \frac{\text{opposite}}{\text{hypotenuse}}

, take opposite

= 3

and hypotenuse

= 5

By Pythagoras theorem,

\text{adjacent}^2 = 5^2 - 3^2 = 25 - 9 = 16

So the adjacent side is

4

Thus,

\tan x = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4}

and

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

Now evaluate

\tan x + 2\cos x = \frac{3}{4} + 2\left(\frac{4}{5}\right) = \frac{3}{4} + \frac{8}{5}

Using a common denominator of

20

, we get

\frac{15}{20} + \frac{32}{20} = \frac{47}{20}

Answer:

\frac{47}{20}

which equals

2\frac{7}{20}

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