Question

In Fig. 9.24, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Answer 1

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 $∆POR,$
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)
$∠OPR + 160° = 180° 2$
$∠OPR = 180° − 160° = 20º 2$
$∠OPR = 10°$