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
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 70° is correct.