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°


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