Answer is B. 70°
Given:
∠POQ = 110°
Property 1: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.
Property 2: Sum of all angles of a quadrilateral = 360°.
By property 1,
∠TPO = 90°
∠TQO = 90°
By property 2,
∠POQ + ∠ TPO + ∠ TQO + ∠PTQ = 360°
⇒ ∠PTQ = 360° - ∠POQ + ∠ TPO + ∠ TQO
⇒ ∠PTQ = 360° - (110° + 90° + 90°)
⇒ ∠PTQ = 360° - 290°
⇒ ∠PTQ = 70°
Hence, ∠PTQ = 70°