JAMB 1984 · UME · Q46

Circle geometry

In the figure, $O$ is the centre of circle $PQRS$ , $PR$ and $SQ$ are diameters, and $PS \parallel RT$ . If $\angle PRT = 135^\circ$ , find $\angle PSQ$ .

Worked solution (try it first)

$PS \parallel RT$ , so co-interior angles add up to $180^\circ$ : $\angle SPR = 180^\circ - 135^\circ$
$= 45^\circ$ .
$OP = OS$ (radii), so triangle $OPS$ is isosceles and $\angle OSP = \angle OPS = 45^\circ$ .
$SQ$ is a diameter through $O$ , so $\angle PSQ = \angle PSO = 45^\circ$ , option B.

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