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(A) \( 60° \)

(B) \( 70° \)

(C) \( 80° \)

(D) \( 90° \)

Answer :

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.

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