In Figure, if TP and TQ are the two tangents to a circle with centre O such that ∠POQ = 110°, then ∠PTQ is equal to. 70° 90° 80° 60°

The correct option is A 70°
Join PQ in given figure then we have two triangles ΔOPQ and ΔPTQ,
In ΔOPQ , OP = OQ (radius of triangle), and in ΔPTQ, PT = TQ (Property of circle, two tangents drawn from a point to a circle are equal), so Both triangles are isosceles triangle
Now in ΔOPQ, by Angle sum property of a triangle
∠OPQ + ∠OQP + ∠POQ = 180°
∠OPQ = ∠OQP = 35°
Since PT is tangent ∠OPT = 90°
∠QPT = 90° - 35° = 55° = ∠PQT
Now ΔPTQ, ∠PTQ + ∠TPQ + ∠TQP = 180°
∠PTQ = 180° - 110° = 70°