Question

0 is the centre of a circle. Tangent TP & TQ of the circle intersect at point T in the

exterior of the circle. Point P & Q lie on the circle. IfPOQ = 1200

then PTQ = ?​

Answer 1

Answer:

Given, ∠POQ=110

∘

We know,

∠OPT=∠OQT=90

∘

(Angle between the tangent and the radial line at the point of intersection of the tangent at the circle)

Now, in quadrilateral POQT

Sum of angles=360

∠OPT+∠OQT+∠PTQ+∠POQ=360

∘

90+90+∠PTQ+110=360

∠PTQ=360−290

∠PTQ=70

∘

hope it's helpful