Question

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

• A. 20°
• B. 10°
• C. 30°
• D. 25°

Answer 1

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

Take a point $S$ in the major arc. Join $PS$ and $RS.$

$PQRS$ is a cyclic quadrilateral.

The sum of either pair of opposite angles of a cyclic quadrilateral is ${180}^{\circ }$.

$\angle PQR+\angle PSR={180}^{\circ }$

$\Rightarrow {100}^{\circ }+\angle PSR={180}^{\circ }$

$\Rightarrow \angle PSR={180}^{\circ }-{100}^{\circ }$

$\Rightarrow \angle PSR={80}^{\circ }\dots \dots \left(i\right)$

The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

Now, $\angle POR=2\angle PSR$

$\Rightarrow \angle POR=2{80}^{\circ }={160}^{\circ }\dots \dots \left(ii\right)$ [Using $eq.\left(i\right)$]

In $△OPR,$ we have

$\because OP=OR$ [radii of a circle]

$\therefore \angle OPR=\angle ORP\dots \dots \left(iii\right)$ [Angles opposite to equal sides of a triangle are equal]

In$△OPR,$ we have

$\angle OPR+\angle ORP+\angle POR={180}^{\circ }$ [Sum of all the angles of a triangle is ${180}^{\circ }$]

$2\angle OPR+\angle OPR+{160}^{\circ }={180}^{\circ }$ [Using eq.(ii) and eq.(iii)]

$\Rightarrow 2\angle OPR+{160}^{\circ }={180}^{\circ }$

$\Rightarrow 2\angle OPR={180}^{\circ }-{160}^{\circ }={20}^{\circ }$

$\Rightarrow \angle OPR={10}^{\circ }$