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c is the correct answer
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Given- O is the centre of a circle to which a pair of tangents PQ&PR from a point P touch the circle at Q&R respectively. ∠RPQ=60
o
.
To find out- ∠ROQ=?
Solution- ∠OQP=90
o
=∠ORP since the angle, between a tangent to a circle and the radius of the same circle passing through the point of contact, is 90
o
. ∴ By angle sum property of quadrilaterals, we get ∠OQP+∠RPQ+∠ORP+∠ROQ=360
o
⟹90
o
+60
o
+90
o
+∠ROQ=360
o
⟹∠ROQ=120
o
.
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