Question

In the figure, $\angle PQR={100}^{\circ },$ where P, Q and R are points on a circle with centre O. Find $\angle OPR.$

• A. ${20}^{\circ }$
• B. ${10}^{\circ }$
• C. ${30}^{\circ }$
• D. ${15}^{\circ }$

Answer 1

The correct option is B ${10}^{\circ }$


Consider the cyclic quadrilateral PQRM.
Since opposite angles are supplementary in a cyclic quadrilateral, we have
$\angle PQR+\angle PMR={180}^{\circ }$
$\therefore \angle PMR={180}^{\circ }-\angle PQR={180}^{\circ }-{100}^{\circ }={80}^{\circ }.$

Since the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the remaining part of the circle and since $\angle PMR={80}^{\circ },$ we have
$\angle POR=2\times \angle PMR=2\times {80}^{\circ }={160}^{\circ }.$

Now, consider the $△OPR,$ which is isosceles as OP=OR=radii of the circle.
$⟹\angle OPR=\angle ORP$
But by angle sum property, we have
$\angle OPR+\angle ORP+\angle POR={180}^{\circ }.$
$⟹2\angle OPR+\angle POR={180}^{\circ }$
$⟹2\angle OPR={180}^{\circ }-\angle POR={180}^{\circ }-{160}^{\circ }={20}^{\circ }$
$\therefore \angle OPR={10}^{\circ }$