WAEC 2023 · Paper 1 · Q37

In the diagram, $O$ is the centre of the circle and $\angle SQR = 28^\circ$ . Find $\angle ORS$ .

Worked solution (try it first)

$\angle SOR$ is at the centre on the same arc $SR$ as $\angle SQR$ .
So $\angle SOR = 2 \times 28^\circ = 56^\circ$ .
$OS = OR$ (radii), so triangle $SOR$ is isosceles: $\angle ORS = \dfrac{180^\circ - 56^\circ}{2}$
$= 62^\circ$ , option B.