JAMB 1990 · UME · Q33

In the figure, $PS = QS = RS$ and $\angle QSR = 100^\circ$ . Find $\angle QPR$ .

Worked solution (try it first)

$SP = SQ = SR$ , so $P$ , $Q$ and $R$ lie on a circle with centre $S$ .
$\angle QSR$ is at the centre and $\angle QPR$ is at the circumference, both on arc $QR$ .
So $\angle QPR = \frac12 \times 100^\circ$
$= 50^\circ$ , option B.