Question

TP and TQ are tangents rlto a circle with centre O, touches the circle at P and Q. if angle POQ 110° then find angle 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

∘

Answer 2

Answer:

70

Step-by-step explanation:

Given, ∠POQ=110∘

 

Since  ∠OPT=∠OQT=90∘

 (Angle between the tangent and the radius line at the point of intersection of the tangent at the circle is always a right angle)

Since in a cyclic quadrilateral POQT  Sum of angles=360

∠OPT+∠OQT+∠PTQ+∠POQ=360∘

 

90+90+∠PTQ+110=360

∠PTQ=360−290

∠PTQ=70∘