JAMB 2003 · UME · Q23

In the diagram, $O$ is the centre of the circle, $POM$ is a diameter and $\angle MNQ = 42^\circ$ . Calculate $\angle QMP$ .

Worked solution (try it first)

Angles in the same segment are equal.
$\angle MPQ$ and $\angle MNQ$ both stand on arc $MQ$ , so $\angle MPQ = 42^\circ$ .
$PM$ is a diameter, so $\angle MQP = 90^\circ$ (angle in a semicircle).
Triangle $QMP$ : $\angle QMP = 180^\circ - 90^\circ - 42^\circ$
$= 48^\circ$ , option D.