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.
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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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