JAMB 1998 · UME · Q25

Circle geometry

In the diagram, $PR$ is a diameter of the circle $PQRS$ , and $PST$ and $QRT$ are straight lines. If $\angle SPR = 35^\circ$ and $\angle PTQ = 30^\circ$ , find $\angle QSR$ .

Worked solution (try it first)

$PR$ is a diameter, so $\angle PSR = 90^\circ$ .
$PST$ is a straight line, so $\angle RST = 90^\circ$ too.
Triangle $RST$ : $\angle SRT = 180^\circ - 90^\circ - 30^\circ$
$= 60^\circ$ .
$QRT$ is straight, so $\angle QRS = 180^\circ - 60^\circ$
$= 120^\circ$ .
Opposite angles of cyclic quadrilateral $PQRS$ add up to $180^\circ$ , so $\angle QPS = 60^\circ$ and $\angle QPR = 60^\circ - 35^\circ$
$= 25^\circ$ .
$\angle QSR$ and $\angle QPR$ both stand on arc $QR$ , so $\angle QSR = 25^\circ$ , option B.

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