JAMB 1993 · UME · Q29

In the diagram, $O$ is the centre of the circle and $POQ$ is a diameter. If $\angle POR = 96^\circ$ , find the value of $\angle ORQ$ .

Worked solution (try it first)

$POQ$ is a straight line, so $\angle QOR = 180^\circ - 96^\circ$
$= 84^\circ$ .
$OQ = OR$ (radii), so triangle $OQR$ is isosceles and its base angles at $Q$ and $R$ are equal.
So $\angle ORQ = \dfrac{180^\circ - 84^\circ}{2}$
$= 48^\circ$ , option B.