In the given figure, O is the centre of a circle PT and PQ are tangents to the circle from an external point P.

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.

$\therefore \angle \mathrm{TRQ}=\frac{1}{2}(\angle \mathrm{QOT})=55^{\circ}$