JAMB 2001 · UME · Q26

A point $P$ moves such that it is equidistant from the points $Q$ and $R$ . Find $QR$ when $PR = 8$ cm and $\angle PRQ = 30^\circ$ .

Worked solution (try it first)

$P$ is equidistant from $Q$ and $R$ , so $PQ = PR = 8$ cm and triangle $PQR$ is isosceles with $\angle PQR = \angle PRQ = 30^\circ$ .
The perpendicular from $P$ meets $QR$ at its mid-point.
Half of $QR$ is $8\cos30^\circ = 8 \times \frac{\sqrt3}{2}$
$= 4\sqrt3$ cm.
So $QR = 2 \times 4\sqrt3 = 8\sqrt3$ cm, option D.