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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