WAEC 2020 · Paper 1 · Q36

In the diagram, $O$ is the centre of the circle. $SOQ$ is the diameter and $\angle SRP = 37^\circ$ . Find $\angle PSQ$ .

Worked solution (try it first)

$\angle SQP$ and $\angle SRP$ stand on the same arc $SP$ , so they are equal: $\angle SQP = 37^\circ$ .
$SQ$ is a diameter, so the angle in the semicircle is a right angle: $\angle SPQ = 90^\circ$ .
The angles of triangle $SPQ$ add up to $180^\circ$ : $\angle PSQ = 180^\circ - 90^\circ - 37^\circ$
$= 53^\circ$ , option C.