WAEC Mathematics 2014 Paper

Concept:Use the tangent-radius property and the isosceles triangle formed by the two radii and the chord.Explanation:Since $PR$ is a tangent to the circle at $Q$, the radius $OQ$ is perpendicular to $PR$.Therefore, $\angle OQR = 90^\circ$.In $\triangle OQS$, $OQ = OS$ because both are radii of the same circle.So, $\triangle OQS$ is isosceles, and $\angle OQS = \angle QSO$.The sum of angles in $\triangle OQS$ is $180^\circ$, and $\angle SOQ = 86^\circ$.Hence, $2\angle OQS + 86^\circ = 180^\circ$.This gives $\angle OQS = \frac{180^\circ - 86^\circ}{2} = 47^\circ$.In the diagram, $QS$ lies between $QO$ and $QR$, so$\angle SQR = \angle OQR - \angle OQS = 90^\circ - 47^\circ = 43^\circ$.Answer:$\angle SQR = 43^\circ$, which is Option A.