In Fig. 10.37, ∠ PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠ OPR.

Consider PR as a chord of the circle. Take any point S on the major arc of the circle

PQRS is a cyclic quadrilateral

∠PQR + ∠PSR = 180° (Opposite angles of a cyclic quadrilateral)

∠PSR = 180° − 100° = 80°

We know that 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

∠POR = 2 ∠PSR

= 2 (80°)

= 160°

In ΔPOR,

OP = OR (Radii of the same circle)

∠OPR = ∠ORP (Angles opposite to equal sides of a triangle)

∠OPR + ∠ORP + ∠POR = 180° (Angle sum property of a triangle)

∠OPR + 160° = 180°

2 ∠OPR = 180° − 160°

2 ∠OPR = 20°

∠OPR = 10°