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

Given:∠PQR=100∘∠POR=2×∠PQR=2×100°=200°∴∠POR=360°𕒶00°=160°InΔOPR,⇒OP=OR...Radii of the circle⇒∠OPR=∠ORPNow,∠OPR+∠ORP+∠POR=180°...Sum of the angles in a triangle⇒∠OPR+∠OPR+160°=180°⇒2∠OPR=180°�°⇒∠OPR=10°