Question

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

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

Answer 1

The correct option is A ${10}^o$

Take a point S in the major arc. Join PS and RS.
$\because$ PQRS is a cyclic quadrilateral.
$\therefore$ $\angle PQR+\angle PSR={180}^o$
[The sum of either pair of opposite angles of a cyclic quadrilateral is ${180}^o$]
$\Rightarrow {100}^o+\angle PSR={180}^o\Rightarrow \angle PSR={180}^o-{100}^o$
$\Rightarrow \angle PSR={80}^o$ ....(i)
Now, $\angle POR=2\angle PSR$
[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]
= $2\times {80}^o={160}^o$ ....(2) [Using (i)]
In $\angle OPR$,
$\because OP=OR$ [radii of a circle]
$therefore\angle OPR=\angle ORP$ ....(3)
[Angles opposite to equal sides of a triangle is 180^{o}]
In $\Delta OPR$,
$\angle OPR+\angle ORP+\angle POR={180}^o$
[Sum of all the angles of a triangle is ${180}^o$]
$\Rightarrow \angle OPR+\angle ORP+{160}^o={180}^o$ [Using (2) and (1)]
$\Rightarrow 2\angle OPR+{160}^o={180}^o$
$\Rightarrow 2\angle OPR={180}^o-{160}^o={20}^o$
$\angle OPR={10}^o$