In Fig. if TP and TQ are tangents drawn from an external point T to a

1.	In Fig. if TP and TQ are tangents drawn from an external point T to a circle with centre O such that ∠TQP = 60°, then ∠OPQ =A. 25° B. 30° C. 40° D. 60°
Answer» 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°

In Fig. if TP and TQ are tangents drawn from an external point T to a circle with centre O such that ∠TQP = 60°, then ∠OPQ =A. 25° B. 30° C. 40° D. 60°

Answer»

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°

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