Test Index

WAEC Mathematics 2021 Paper

© africaexams.com

Question : 19

Total: 49

In △LMN, |LM| = 6cm, ∠LNM = x and sin x = sin x = $\; \frac{3}{5}$ .

60cm $^{2}$
48cm $^{2}$
24cm $^{2}$
30cm $^{2}$

Solution:

Concept:

Use

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

and the area of a right-angled triangle:

\frac12 \times \text{base} \times \text{height}

.

Explanation:

In

\triangle LMN

,

\angle LNM = x

, and

LM = 6

cm is opposite this angle.

Given

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

, and

\sin x = \frac{6}{\text{hypotenuse}}

.

So,

\frac{6}{\text{hypotenuse}} = \frac{3}{5}

.

Therefore, the hypotenuse

= 10

cm.

Using Pythagoras, the other leg

= \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = 8

cm.

Area of the triangle

= \frac12 \times 6 \times 8 = 24

cm

^2

.

Answer:

24

cm

^2

— Option C.

© africaexams.com

Go to Question:

Prev Question

Next Question