In Fig. 10.11, 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°

Given: TP and TQ are tangents.

Therefore, radius drawn to these tangents from centre of the circle will be perpendicular to the tangents.

Thus, OP ⊥ TP and OQ ⊥ TQ

And therefore,

∠OPT = 90°

∠OQT = 90°

In quadrilateral POQT,

Sum of all interior angles = 360°

∠OPT + ∠POQ +∠OQT + ∠PTQ = 360°

90°+ 110° + 90° + ∠PTQ = 360°

∠PTQ = 70°