JAMB 1993 · UME · Q28

In the diagram, $PQRS$ is a circle with centre $O$ and diameter $SQ$ , and $PQ \parallel RT$ . If $\angle RTS = 32^\circ$ , find $\angle PSQ$ .

Worked solution (try it first)

$PQ \parallel RT$ and $SQT$ is a straight line, so alternate angles are equal: $\angle PQS = \angle RTS = 32^\circ$ .
$SQ$ is a diameter, so $\angle SPQ = 90^\circ$ (angle in a semicircle).
The angles of triangle $PSQ$ add up to $180^\circ$ : $\angle PSQ = 180^\circ - 90^\circ - 32^\circ$
$= 58^\circ$ , option C.