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°