In the given figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110, then ∠PTQ is equal to
(A) 60 (B) 70
(C) 80 (D) 90
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 = 360
∠OPT + ∠POQ +∠OQT + ∠PTQ = 360
⇒ 90+ 110º + 90 +PTQ = 360
⇒ PTQ = 70
Hence, alternative (B) is correct.