JAMB 1999 · UME · Q31

Circle geometry

In the figure, $PQRS$ is a circle with $ST \parallel RQ$ ( $T$ on $PQ$ ) and $PT = PS$ . If $\angle QRS = 110^\circ$ , find the value of $x = \angle PQR$ .

Worked solution (try it first)

Opposite angles of cyclic quadrilateral $PQRS$ add up to $180^\circ$ : $\angle QPS = 180^\circ - 110^\circ$
$= 70^\circ$ .
$PT = PS$ , so triangle $PTS$ is isosceles: $\angle PTS = \dfrac{180^\circ - 70^\circ}{2}$
$= 55^\circ$ .
$ST \parallel RQ$ , so corresponding angles are equal: $x = \angle PTS$ , which is $55^\circ$ , option B.

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