JAMB 2010 · UTME · Q39

If the area of $\triangle PQR$ is $12\sqrt3\text{ cm}^2$ , $|PQ| = 8\text{ cm}$ , $|PR| = q$ and $\angle QPR = 60^\circ$ , find the value of $q$ .

Worked solution (try it first)

Use area $= \frac12 ab\sin C$ with the sides $PQ = 8$ and $PR = q$ and the angle $60^\circ$ between them.
Area $= \frac12 \times 8 \times q \times \frac{\sqrt3}{2}$
$= 2\sqrt3\,q$ .
Set $2\sqrt3\,q = 12\sqrt3$ and divide both sides by $2\sqrt3$ : $q = 6$ cm, option A.