JAMB 2002 · UME · Q14

In the diagram, $PST$ is a straight line and $PQ = QS = RS$ . If $\angle RST = 72^\circ$ , find $x$ .

Worked solution (try it first)

$PQ = QS$ , so $\angle QSP = \angle QPS = x$ .
The exterior angle of triangle $PQS$ at $Q$ is $\angle SQR = x + x = 2x$ .
$QS = RS$ , so $\angle QRS = \angle SQR = 2x$ .
$\angle RST$ is an exterior angle of triangle $PRS$ , so it equals $\angle P + \angle R$ : $x + 2x = 72^\circ$ .
So $3x = 72^\circ$ and $x = 24^\circ$ , option C.