JAMB 2004 · UME · Q26

$P$ , $R$ and $S$ lie on a circle centre $O$ as shown, while $Q$ lies outside the circle. Find $\angle PSO$ .

Worked solution (try it first)

$Q$ , $R$ and $S$ are in a straight line, so $\angle PRS$ is the exterior angle of triangle $PQR$ : $\angle PRS = 20^\circ + 35^\circ$
$= 55^\circ$ .
The angle at the centre is twice the angle at the circumference on the same arc $PS$ : $\angle POS = 2 \times 55^\circ$
$= 110^\circ$ .
$OP = OS$ (radii), so triangle $POS$ is isosceles and its base angles are equal.
So $\angle PSO = \frac{180^\circ - 110^\circ}{2}$
$= 35^\circ$ , option A.