JAMB 1991 · UME · Q38

In the figure, $PQ = PR = PS$ , $QR$ is produced to $T$ and $\angle SRT = 68^\circ$ . Find $\angle QPS$ .

Worked solution (try it first)

$PQ = PR = PS$ , so $Q$ , $R$ and $S$ lie on a circle with centre $P$ .
Angles on a straight line: $\angle QRS = 180^\circ - 68^\circ$
$= 112^\circ$ .
It stands on the major arc $QS$ , so the reflex angle at $P$ is $2 \times 112^\circ = 224^\circ$ .
Angles at a point add up to $360^\circ$ : $\angle QPS = 360^\circ - 224^\circ$
$= 136^\circ$ , option A.