In figure displayed below, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ=110∘, then ∠PTQ is equal to
OP and OQ are radii of the circle to the tangents TP and TQ respectively.
Since, the line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
∴ OP ⊥ TP and OQ ⊥ TQ.
∠OPT=∠OQT=90∘
In quadrilateral POQT,
Sum of all interior angles = 360∘
∠PTQ+∠OPT+∠POQ+∠OQT=360∘⇒∠PTQ+90∘+110∘+90∘=360∘⇒∠PTQ=70∘
∠PTQ is equal to option (A) 70∘.