Question

In the figure if TP, Tq tangents drawn an external point T to a circle with
center 'o' such that < TQP = 60, then find <OPQ.​

Answer 1

Given:

✰ TP and TQ are tangents drawn from an external point T to a circle with center O.

✰ ∠TQP = 60°

To find:

✠ ∠OPQ

Solution:

In the given figure TP and TQ are the tangents drawn from an external point T to a circle with center O.

OP, OQ and PQ are joined

∠TQP = 60°

In ∆TPQ,

∵ TP = TQ [ The tangents drawn from an external point T to a circle ]

∴ ∠TQP = ∠TPQ = 60°. [ ∵ TP = TQ ]

∴ ∠PTQ = 180° - ( 60° + 60° )

⟹ ∠PTQ = 180° - 120°

⟹ ∠PTQ = 60°

Then,

⟹ ∠POQ = 180° - ∠PTQ

⟹ ∠POQ = 180° - 60°

⟹ ∠POQ = 120°

Now,

In ∆POQ

⟹ OP = OQ. [ ∵ OP and OQ are the radii of the same circle ]

⟹ ∠OPQ = ∠OQP

then,

⟹ ∠OPQ + ∠OQP = 180° - ∠POQ

⟹ ∠OPQ + ∠OQP = 180° - 120°

⟹ ∠OPQ + ∠OQP = 60°

➛∠OPQ + ∠OPQ = 60°. [ ∵ ∠OPQ = ∠OQP ]

➛ 2∠OPQ = 60°

➛ ∠OPQ = 60°/2

➛ ∠OPQ = 30°

∴ ∠OPQ = 30°

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