In the figure,angle PQR = 100^{circ},where P, | vedclass

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(B) The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at any point on the remaining part of the cir

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$10$

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(B) The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at any point on the remaining part of the circle.Therefore,reflex $\angle POR = 2 	imes \angle PQR$.Given $\angle PQR = 100^{\circ}$.So,reflex $\angle POR = 2 	imes 100^{\circ} = 200^{\circ}$.Since $\angle POR + 	ext{reflex } \angle POR = 360^{\circ}$,we have $\angle POR + 200^{\circ} = 360^{\circ}$.Thus,$\angle POR = 360^{\circ} - 200^{\circ} = 160^{\circ}$.In $\Delta POR$,$OP = OR$ (radii of the same circle).Therefore,$\angle OPR = \angle ORP$ (angles opposite to equal sides are equal).In $\Delta POR$,the sum of angles is $180^{\circ}$,so $\angle OPR + \angle ORP + \angle POR = 180^{\circ}$.Substituting $\angle ORP = \angle OPR$,we get $2 \angle OPR + 160^{\circ} = 180^{\circ}$.$2 \angle OPR = 180^{\circ} - 160^{\circ} = 20^{\circ}$.$\angle OPR = \frac{20^{\circ}}{2} = 10^{\circ}$.

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

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