Question

4(a). In the diagram, P, Q, R and S are points on the circle centre K. KR is a bisector of ∠SRQ, ∠KSP=41° and ∠SKR=80°. Find: (a) ∠RQP; (b) ∠SPQ.

Accepted Answer

(a) Since triangle KSR is isosceles (KS=KR, radii) with base angles equal: ∠KSR=∠KRS. 2∠KSR+80°=180° → ∠KSR=50°. Since ∠ROP+∠RSP=180° (opposite angles of a cyclic quadrilateral): ∠RQP+∠RSP=180°. ∠RSP=∠RSK+∠KSP=50°+41°=91°. So ∠RQP=180°-91°=89°.
(b) ∠SRQ=2×∠KRS=2×50°=100° (since KR bisects ∠SRQ). ∠SRQ+∠SPQ=180° (opposite angles of a cyclic quadrilateral): 100°+∠SPQ=180° → ∠SPQ=80°.

WAEC Mathematics 2023 Theory — Question 4

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