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°



(B)

Given,


TP and TQ are tangents.


Therefore, radius drawn to these tangents will be perpendicular to the tangents.


Thus, OP TP and OQ TQ


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°


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