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In the adjoining figure, O is the centre of the circle and P, Q and R are points on the circle such that $∠ PQR = 100 °,$ then $∠ OPR$ equals:

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80 °

10 °

100 °

60 °

answer is B.

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Detailed Solution

Given that, O is the centre of the circle and P, Q and Rare points on the circle such that

∠ PQR = 100 ° .

Here, PR is chord we marks on major arc of the circle.

∴

PQRS is a cyclic quadrilateral.
So, the sum of opposite angles of cyclic quadrilateral PQRS is

180 °

∠ P Q R + ∠ P S R = 180 °

⇒ 100 + ∠ P S R = 180 °

⇒ ∠ P S R = 180 ° − 100 °

⇒ ∠ P S R = 80 °

We have to find the

∠ O P R

,
Here, the arc PQR subtends

∠ P Q R

at centre of a circle and

∠ P S R

on points.
So, the angle subtended by arc PQR at the centre is double the angle subtended by it at any other point on the circle.

⇒ ∠ P O R = 2 ∠ P S R

⇒ ∠ P O R = 2 × 80 ° ⇒ ∠ P O R = 160 °

Now,
In

Δ

OPR, OP = OR (Radii of same circle arc equals)

∴ ∠ O P R

∠ O R P

(Opposite angles to equal sides are equals) … (1)
Also, in

Δ O P R

∠ O P R + ∠ O R P + ∠ P O R = 180 °

(Angle sum property of triangle)

⇒ ∠ O P R + ∠ O P R + ∠ P O R = 180 °

From (1),

⇒ 2 ∠ O P R + 160 ° = 180 °

⇒ 2 ∠ O P R = 180 ° − 160 °

⇒ 2 ∠ O P R = 20 °

⇒ ∠ O P R = 202

⇒ ∠ O P R = 10 °

The value of

∠ O P R = 10 °

.
Hence, the correct option is 2.

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