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^∘ then ∠TRQ
[CBSE 2015]

Answer 1

Construction: Join OQ and OT


We know that the radius and tangent are perperpendular at their point of contact
∵∠OTP = ∠OQP = 90^∘
Now, In quadrilateral OQPT
∠QOT + ∠OTP + ∠OQP + ∠TPQ = 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 centre is double the angle subtended by the arc at any point on the remaining part of the circle.
$\therefore \angle \mathrm{TRQ}=\frac{1}{2}\left(\angle \mathrm{QOT}\right)=55°$