JAMB 2000 · UME · Q29

In the diagram, $\angle RPS = 50^\circ$ , $\angle RPQ = 30^\circ$ and $PQ = QR$ . Find the value of $\angle PRS$ .

Worked solution (try it first)

$PQ = QR$ , so triangle $PQR$ is isosceles and $\angle QRP = \angle QPR = 30^\circ$ .
So $\angle PQR = 180^\circ - 60^\circ$
$= 120^\circ$ .
Opposite angles of cyclic quadrilateral $PQRS$ add up to $180^\circ$ : $\angle PSR = 180^\circ - 120^\circ$
$= 60^\circ$ .
Triangle $PRS$ : $\angle PRS = 180^\circ - 50^\circ - 60^\circ$
$= 70^\circ$ , option B.