WAEC 2017 · Paper 2 · Q4

(a)
The diagram shows a circle centre $O$ and $\overline{PQ}$ is a diameter. $S$ and $R$ are points on the circle with $|SR| = |RQ|$ . If $\angle PSO = 44^\circ$ , calculate the value of $\angle OQR$ .

Worked solution (try it first)

$OS = OP$ (radii), so triangle $OSP$ is isosceles and $\angle SPO = \angle PSO = 44^\circ$ .
$\angle SOQ$ is an exterior angle of triangle $OSP$ : $\angle SOQ = 44^\circ + 44^\circ$
$= 88^\circ$ .
$R$ is on the minor arc $SQ$ , so $\angle SRQ$ stands on the major arc.
The reflex angle $SOQ$ is $360^\circ - 88^\circ = 272^\circ$ , and $\angle SRQ = \frac12 \times 272^\circ$
$= 136^\circ$ .
$OS = OQ$ and $RS = RQ$ , so $OSRQ$ is a kite, and $\angle OSR = \angle OQR$ .
The angles of the quadrilateral add up to $360^\circ$ : $88^\circ + 136^\circ + 2\angle OQR = 360^\circ$ .
So $2\angle OQR = 136^\circ$ and $\angle OQR = 68^\circ$ .