Question

O is the centre of the circle. PQ is a chord and PT is is tangent to circle at P. If angle POQ=70degrees, find angle TPQ

Answer 1

Answer:

∠TPQ = 35°

For better understanding of the solution, see the attached figure of the problem :

Since, OP and OQ both are radius of same circle

⇒ OP = OQ

Therefore, ΔOPQ is an isosceles triangle since two sides are equal

Now, by using property of isosceles triangle that the corresponding angles to equal sides in an isosceles triangle are equal. We get,

∠OPQ = ∠OQP

By using Angle sum property of triangle in ΔOPQ :

∠POQ + ∠OPQ + ∠OQP = 180°

⇒ 70° + 2∠OPQ = 180°

⇒ 2∠OPQ = 110°

⇒ ∠OPQ = 55°

Now, tangent makes right angles with the point of contact with the circle.

⇒ ∠OPT = 90°

⇒ ∠OPQ + ∠TPQ = 90°

⇒ 55° + ∠TPQ = 90°

⇒ ∠TPQ = 35°