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JAMB Mathematics 2014 Paper

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Question : 44

Total: 47

Find the value of x in the figure above

20 $\sqrt{3}$ cm
10 $\sqrt{3}$ cm
5 $\sqrt{3}$ cm
4 $\sqrt{3}$ cm

Solution:

Concept:

Use the sine rule:

\frac{a}{\sin A} = \frac{b}{\sin B}

for any triangle.

Explanation:

In the triangle, side

x

is opposite

60^\circ

.

The side of length

10

cm is opposite

30^\circ

.

Applying the sine rule:

\frac{x}{\sin 60^\circ} = \frac{10}{\sin 30^\circ}

.

So

x = \frac{10 \sin 60^\circ}{\sin 30^\circ}

.

We know

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

and

\sin 30^\circ = \frac{1}{2}

.

Thus

x = \frac{10 \times \frac{\sqrt{3}}{2}}{\frac{1}{2}} = 10 \times \frac{\sqrt{3}}{2} \times \frac{2}{1} = 10\sqrt{3}

cm.

Answer:

x = 10\sqrt{3}

cm, which is option B.

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