WAEC 2023 · Paper 2 · Q4

Circle geometry

In the diagram, $P$ , $Q$ , $R$ and $S$ are points on the circle centre $K$ . $\overline{KR}$ is a bisector of $\angle SRQ$ , $\angle KSP = 41^\circ$ and $\angle SKR = 80^\circ$ . Find:

(a)
$\angle RQP$ ;
(b)
$\angle SPQ$ .

Worked solution (try it first)

(a)

$KR = KS$ (radii), so triangle $KRS$ is isosceles and $\angle KRS = \angle KSR$
$= \frac{180^\circ - 80^\circ}{2}$
$= 50^\circ$ .
So $\angle RSP = \angle RSK + \angle KSP$
$= 50^\circ + 41^\circ$
$= 91^\circ$ .
$PQRS$ is a cyclic quadrilateral, so $\angle RQP = 180^\circ - 91^\circ$
$= 89^\circ$ .

(b)

$KR$ bisects $\angle SRQ$ , so $\angle SRQ = 2 \times 50^\circ$
$= 100^\circ$ .
Opposite angles of the cyclic quadrilateral: $\angle SPQ = 180^\circ - 100^\circ$
$= 80^\circ$ .

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