Answer is B. 45°
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, ∆POQ is right-angled at ∠OPQ (i.e., ∠OPQ = 90°).
∵ ∆POQ is an isosceles triangle
∴ ∠POQ = ∠OQP
By property 2,
∠POQ + ∠OQP + ∠QPO = 180°
⇒ ∠POQ + ∠OQP = 180° - ∠QPO
⇒ ∠POQ + ∠OQP = 180° - 90°
⇒ ∠POQ + ∠OQP = 180° - 90°
⇒ ∠POQ + ∠OQP = 90°
⇒ ∠OQP + ∠OQP = 90° [∵∠POQ = ∠OQP]
⇒ 2∠OQP = 90°
⇒ ∠OQP = \(\frac{90^°}{2}\) ⇒ ∠OQP = 45°
Hence, ∠OQP = 45°