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°