WAEC 2015 · Paper 2 · Q3

(a)
The ratio of the interior angle to the exterior angle of a regular polygon is $5 : 2$ . Find the number of sides of the polygon.
(b)
The diagram shows a circle $PQRS$ with centre $O$ ; $TPQU$ is a straight line. $\angle UQR = 68^\circ$ , $\angle TPS = 74^\circ$ and $\angle QSR = 40^\circ$ . Calculate the value of $\angle PRS$ .

Worked solution (try it first)

At each corner, interior angle + exterior angle $= 180^\circ$ .
Sharing $180^\circ$ in the ratio $5 : 2$ (7 parts), the exterior angle is $\frac27 \times 180^\circ = \frac{360^\circ}{7}$ .
The number of sides is $360^\circ$ divided by the exterior angle: $360 \div \frac{360}{7} = 7$ sides.

$\angle QPR$ and $\angle QSR$ stand on the same arc $QR$ , so $\angle QPR = \angle QSR = 40^\circ$ (angles in the same segment).
$\angle UQR = 68^\circ$ is an exterior angle of triangle $PQR$ (as $PQU$ is a straight line), so it equals the two opposite interior angles: $68^\circ = 40^\circ + \angle PRQ$ , giving $\angle PRQ = 28^\circ$ .
$\angle TPS = 74^\circ$ is an exterior angle of the cyclic quadrilateral $PQRS$ , so it equals the interior opposite angle: $\angle SRQ = 74^\circ$ .
So $\angle PRS = \angle SRQ - \angle PRQ$
$= 74^\circ - 28^\circ$
$= 46^\circ$ .