In the given figure, ∠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°
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)
2 ∠OPR + 160° = 180°
2 ∠OPR = 180° − 160° = 20º
∠OPR = 10°
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