Answer :

**Method 1:**

We know angle subtended at centre by an arc is double the angle subtended

by it at any other point.

Reflex angle ∠POR= 2∠PQR

= 2 × 100°

= 200°

Now, ∠POR = 360° – 200 = 160°

**Also,**

PO = OR [Radii of a circle]

∠OPR = ∠ORP [Opposite angles of isosceles triangle]

In ∆OPR, ∠POR = 160°

∴ ∠OPR = ∠ORP = 10°

**Method 2:**

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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*ABCD* is a RD Sharma - Mathematics

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<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

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<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics