JAMB 1990 · UME · Q40

In the figure, $\angle YXZ = 30^\circ$ , $\angle XYZ = 105^\circ$ and $XY = 8$ cm. Calculate $YZ$ .

Worked solution (try it first)

The angles add up to $180^\circ$ , so $\angle Z = 180^\circ - 30^\circ - 105^\circ$
$= 45^\circ$ .
Sine rule: $YZ$ faces the $30^\circ$ at $X$ , and $XY = 8$ faces the $45^\circ$ at $Z$ .
So $\dfrac{YZ}{\sin30^\circ} = \dfrac{8}{\sin45^\circ}$ .
So $YZ = \dfrac{8 \times \frac12}{\frac{\sqrt2}{2}}$
$= \dfrac{8}{\sqrt2}$ , which is $4\sqrt2$ cm, option C.