WAEC Mathematics 2011 Paper

Concept:Apply the angle-at-the-centre circle theorem and the angle sum of a quadrilateral.Explanation:The reflex $\angle PQR = 240^\circ$, so the ordinary angle at $Q$ is $\angle PQR = 360^\circ - 240^\circ = 120^\circ$.The arc $PR$ not containing $Q$ is intercepted by this $120^\circ$ angle, so it measures $2 \times 120^\circ = 240^\circ$.Therefore the remaining arc $PR$ measures $360^\circ - 240^\circ = 120^\circ$, giving the central angle $\angle POR = 120^\circ$.Since $PQ = QR$ and $OP = OR$ are radii, quadrilateral $PQRO$ is symmetric, so the angles at $P$ and $R$ are equal. Let each equal $x$.Using the sum of interior angles of quadrilateral $PQRO$:$x + 120^\circ + x + 120^\circ = 360^\circ$$2x + 240^\circ = 360^\circ$$2x = 120^\circ$$x = 60^\circ$.Answer:$x = 60^\circ$