Question

In the given figure, O is the centre of a circle. PT and PQ are tangents to the circle from an external point P. If ∠TPQ = 70°, find ∠TRQ.

Answer 1

Construction: Join OQ and OT

We know that the radius and tangent are perpendicular at their point of contact
∵ ∠OTP = ∠OQP = 90°
Now, In quadrilateral OQPT
∠QOT + ∠OTP +∠OQP+ ∠TPO = 360°       [Angle sum property of a quadrilateral]
⇒ ∠QOT +90° + 90°+ 70° = 360°
⇒ 250° + ∠QOT =  360°
⇒ ∠ QOT = 110°
We know that the angle subtended by an arc at the center is double the angle subtended by the arc at any point on the remaining part of the circle.
 `∴ ∠ TRQ = 1/2 (∠QOT ) = 55°`