JAMB 1989 · UME · Q37

In $\triangle XYZ$ , $\angle Y = \angle Z = 30^\circ$ and $XZ = 3$ cm. Find $YZ$ .

Worked solution (try it first)

The angles add up to $180^\circ$ , so $\angle X = 180^\circ - 30^\circ - 30^\circ$
$= 120^\circ$ .
Sine rule: $YZ$ faces $\angle X$ and $XZ = 3$ faces $\angle Y$ , so $\dfrac{YZ}{\sin120^\circ} = \dfrac{3}{\sin30^\circ}$ .
So $YZ = \dfrac{3 \times \frac{\sqrt3}{2}}{\frac12}$ , which is $3\sqrt3$ cm, option C.