In Fig. 10.37, PQR = 100°, where P, Q and R are points on a circle with centre O. Find OPR.
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In Fig. 10.37, PQR = 100°, where P, Q and R are points on a circle with centre O. Find OPR.
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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
.