JAMB 1989 · UME · Q34

In the diagram, $PST$ is a straight line, $PQR$ is a straight line and $PQ = QS = SR$ . If $\angle SPQ = 24^\circ$ , find $y = \angle RST$ .

Worked solution (try it first)

$PQ = QS$ , so $\angle QSP = \angle QPS = 24^\circ$ .
The exterior angle of triangle $PQS$ at $Q$ is $\angle SQR = 24^\circ + 24^\circ$
$= 48^\circ$ .
$QS = SR$ , so $\angle SRQ = \angle SQR = 48^\circ$ .
In triangle $QSR$ , $\angle QSR = 180^\circ - 48^\circ - 48^\circ$
$= 84^\circ$ .
$PST$ is a straight line, so $y = 180^\circ - 24^\circ - 84^\circ$
$= 72^\circ$ , option C.