Question

If PT is a tangent to a circle with centre 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 perperpendular at their point of contact.
∴∠OPT = 90^∘
Now, ∠OPQ = ∠OPT − ∠TPQ = 90^∘ − 70^∘ = 20^∘
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 20^∘ (Angles opposite to equal sides are equal)
Now, In isosceles △POQ
∠POQ + ∠OPQ + ∠OQP = 180^∘ (Angle sum property of a triangle)
⇒ ∠POQ = 180^∘ − 20^∘ − 20^∘ = 140^∘