JAMB 1985 · UME · Q40

In the figure, $POQ$ is the diameter of the circle $PQRS$ . If $\angle PSR = 145^\circ$ , find $x^\circ$ (the angle $QPR$ ).

Worked solution (try it first)

$PQRS$ is a cyclic quadrilateral, so opposite angles add up to $180^\circ$ : $\angle PQR = 180^\circ - 145^\circ$
$= 35^\circ$ .
$PQ$ is a diameter, and the angle in a semicircle is $90^\circ$ , so $\angle PRQ = 90^\circ$ .
The angles of triangle $PQR$ add up to $180^\circ$ : $x = 180^\circ - 90^\circ - 35^\circ$
$= 55^\circ$ , option D.