JAMB 1988 · UME · Q30

Circle geometry

In the figure, $P$ , $Q$ , $R$ and $S$ lie on a circle and the chords $QR$ and $RS$ are equal. $SR$ is produced to $T$ with $\angle QRT = 120^\circ$ , and the exterior angle at $S$ (between $SP$ and $RS$ produced) is $100^\circ$ . Calculate the angle $x = \angle PRS$ .

Worked solution (try it first)

Angles on a straight line: $\angle QRS = 180^\circ - 120^\circ$
$= 60^\circ$ .
With $QR = RS$ , triangle $QRS$ is isosceles with a $60^\circ$ apex, so it is equilateral and $\angle SQR = 60^\circ$ .
Angles in the same segment: $\angle SPR$ and $\angle SQR$ both stand on arc $SR$ , so $\angle SPR = 60^\circ$ .
Angles on a straight line at $S$ : $\angle PSR = 180^\circ - 100^\circ$
$= 80^\circ$ .
The angles of triangle $PSR$ add up to $180^\circ$ : $x = 180^\circ - 80^\circ - 60^\circ$
$= 40^\circ$ , option D.

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