JAMB 1984 · UME · Q28

In triangle $XYZ$ , $\angle YXZ = 120^\circ$ , $XZ = 3$ cm and $YZ = 5$ cm. Find $\angle XYZ$ .

Worked solution (try it first)

Use the sine rule.
$YZ = 5$ faces the $120^\circ$ angle at $X$ , and $XZ = 3$ faces $\angle Y$ , so $\dfrac{\sin Y}{3} = \dfrac{\sin120^\circ}{5}$ .
Multiply both sides by 3: $\sin Y = \dfrac{3 \times 0.8660}{5}$
$= 0.5196$ .
So $Y = 31.31^\circ$ .
The decimal part is $0.31 \times 60 \approx 18'$ , so $\angle XYZ = 31^\circ18'$ , option D.