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# In 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) $$90°$$

Since TP and TQ are tangents to a circle with centre O so that $$\angle \ POQ \ = \ 110°$$,

$$\therefore$$$$OP \ ⊥ \ PT$$ and $$OQ \ ⊥ \ QT$$

$$\Rightarrow \ ∠ \ OPT \ = \ 90°$$ and $$∠ \ OQT \ = \ 90°$$

In the quadrilateral TPOQ, we have

$$∠ \ PTQ \ + \ ∠ \ TPO \ + \ ∠ \ POQ \ + \ ∠ \ OQT \ = \ 360°$$
[ \therefore Sum of all angles in a quadrilateral = 360° ]

$$∠ \ PTQ \ + \ 90° \ + \ 110° \ + \ 90° \ = \ 360°$$

$$∠ \ PTQ \ + \ 290° \ = \ 360°$$

$$\Rightarrow \ ∠ \ PTQ \ = \ 360° \ - \ 290°$$

$$\Rightarrow \ ∠ \ PTQ \ = \ 70°$$

$$\therefore$$ Option (B) is correct.