Question

If PT is a tangent to a circle with center O and PQ is a chord of the circle such that ∠QPT =  70° , then find the measure of ∠POQ.

Answer 1

We know that the radius and tangent are perpendicular at their point of contact.

∴ ∠OPT = 90°

Now, ∠OPQ = ∠OPT - ∠TPQ = 90° - 70° = 20°

Since, OP = OQ as both are radius

∴ ∠OPQ = ∠OQP = 20°    (angle opposite to equal sides are equal)

Now, In isosceles △ POQ

∠POQ + ∠OPQ + ∠OQP = 180°  (angle sum property of a triangle)

\(\Rightarrow\) ∠POQ = 180° - 20° = 140°