Answer is B. 30°
Given:
∠TQP = 60°
Property 1: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.
Property 2: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.
By property 1,
TP = TQ (tangent from T)
⇒ ∠TPQ = ∠TQP = 60°
By property 2, ∆OPT is right-angled at ∠OPT (i.e., ∠OPT = 90°) and ∆OQT is right-angled at ∠OQT (i.e., ∠OQT = 90°).
Now,
∠OPQ = ∠OPT – ∠TPQ
⇒ ∠OPQ = 90° – 60°
⇒ ∠OPQ = 30°
Hence, ∠OPQ = 30°
