Question

In the given 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. ${40}^{\circ }$
• C. ${10}^{\circ }$
• D. ${80}^{\circ }$

Answer 1

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


Consider PR as a chord of the circle.
Take any point S on the major arc of the circle. PQRS is a cyclic quadrilateral.
$\angle PQR+\angle PSR={180}^{\circ }\text{(Opposite angles of a cyclic quadrilateral)}$
$\therefore \angle PSR={180}^{\circ }-{100}^{\circ }={80}^{\circ }$
We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
$\Rightarrow \angle POR=2\angle PSR=2\left({80}^{\circ }\right)={160}^{\circ }In\Delta POR,$
OP = OR (Radii of the same circle)
$\therefore \angle OPR=\angle ORP\text{(Angles opposite to equal sides of a triangle)}$
$\angle OPR+\angle ORP+\angle POR={180}^{\circ }\text{(Angle sum property of a triangle)}$
$2\angle OPR+{160}^{\circ }={180}^{\circ }$
$2\angle OPR={180}^{\circ }-{160}^{\circ }={20}^{\circ }$
$\Rightarrow \angle OPR={10}^{\circ }$