Question

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


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Answer 1

ANSWER

Here, PR is chord

We mark s on major arc of the circle.

∴ PQRS is a cyclic quadrilateral.

So, ∠PQR+∠PSR=180

o

[Sum of opposite angles of a cyclic quadrilateral is 180

o

]

100+∠PSR=180

o

∠PSR=180

o

−100

o

∠PSR=80

o

Arc PQR subtends ∠PQR at centre of a circle.

And ∠PSR on point s.

So, ∠POR=2∠PSR

[Angle subtended by arc at the centre is double the angle subtended by it any other point]

∠POR=2×80

o

=160

o

Now,

In ΔOPR,

OP=OR[Radii of same circle are equal]

∴∠OPR=∠ORP [opp. angles to equal sides are equal] ………………..(1)

Also in ΔOPR,

∠OPR+∠ORP+∠POR=180

o

(Angle sum property of triangle)

∠OPR+∠OPR+∠POR=180

o

from (1)

2∠OPR+160=180

o

2∠OPR=180

o

−160

o

2∠OPR=20

∠OPR=20/2

∴∠OPR=10

o

.