**Answer is A.** 100°

Given:

∠QPB =50°

**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 triangle = 180°.

By property 1, ∆OPB is right-angled at ∠OPB (i.e., ∠OPB = 90°).

∠OPQ = ∠OPB – ∠QPB

⇒ ∠OPQ = 90° – 50° = 40°

And,

∠OPQ = ∠OQP [∵ OP = OQ (radius of circle)]

Now by property 2,

∠OPQ + ∠OQP + ∠POQ = 180°

⇒ 40° + 40° + ∠POQ = 180°

⇒ 80° + ∠POQ = 180°

⇒ ∠POQ = 180° - 80°

⇒ ∠POQ = 100°

Hence, ∠POQ = 100°