Question

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

__

Answer 1

∵ PQRS is a cyclic quadrilateral

∴ $\angle$PQR + $\angle$PSR = ${180}^{\circ }$ (sum of either pair of opposite angles of a cyclic quadrilateral is${180}^{\circ }$.

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

$\Rightarrow$ $\angle$PSR = ${80}^{\circ }$ ........ (1)

Now, $\angle$POR subtended = 2$\angle$PSR = ${160}^{\circ }$ ...........(2)

(angle usbtended by an arc at the centre is double the angle by it at any point on the remaining part of circle)

In $\Delta$OPR

∵ OP = OR (radii of circle)

∴ $\angle$OPR = $\angle$ORP ...........(3) (angles opposite to equal sides of a triangle are equal)

In $\Delta$OPR

$\angle$OPR + $\angle$ORP + $\angle$POR = ${180}^{\circ }$

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

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