In Fig, ∠ PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠ OPR.

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Detailed Solution

Mark any point on the major arc side (opposite side to point Q) as X.

Since all points P, Q, R, X lie on the circle, PQRX becomes a cyclic quadrilateral.

We know that the sum of either pair of opposite angles of a cyclic quadrilateral is 180°.

Therefore,

∠PQR + ∠PXR = 180°

100° + ∠PXR = 180°

∠PXR = 180° - 100° = 80°

We know that 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.

Therefore,

∠POR = 2∠PXR

= 2 × 80°

= 160°

Consider the ∆OPR. It is an isosceles triangle as OP = OR = Radius of the circle.

Thus, ∠OPR = ∠ORP

Sum of all angles in a triangle is 180°.

Therefore,

∠OPR + ∠POR + ∠ORP = 180°

∠OPR + 160° + ∠OPR = 180°

2∠OPR = 180° - 160°

∠OPR = 10°

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