Question 2:
In the given figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110º , then ∠PTQ is equal to
(A) 60º (B) 70º (C) 80º (D) 90º
Answer:
It is given that TP and TQ are tangents.
Therefore, radius drawn to these tangents will be perpendicular to the tangents.
Thus, OP ⊥ TP and OQ ⊥ TQ
∠OPT = 90º
∠OQT = 90º
In quadrilateral POQT,
Sum of all interior angles = 360º
∠OPT + ∠POQ +∠OQT + ∠PTQ = 360º
⇒ 90º + 110º + 90º + PTQ = 360º
⇒ PTQ = 70º
Hence, alternative (B) is correct.
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