Question

In the given figure, PQ is a chord of a circle with centre O and PT is a tangent. If ∠QPT = 60^∘ , find ∠PRQ

Answer 1

We know that the radius and tangent are perperpendular at their point of contact.
∴∠OPT = 90^∘
Now, ∠OPQ = ∠OPT − ∠QPT = 90^∘ − 60^∘ = 30^∘
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 30^∘ (Angles opposite to equal sides are equal)
Now, In isosceles △POQ
∠POQ + ∠OPQ + ∠OQP = 180^∘ (Angle sum property of a triangle)
⇒ ∠POQ = 180^∘ − 30^∘ − 30^∘ = 120^∘
Now, ∠POQ + reflex ∠POQ = 360^∘ (Complete angle)
⇒ reflex ∠POQ = 360^∘ − 120^∘ = 240^∘
We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.
$\therefore \angle \mathrm{PRQ}=\frac{1}{2}\left(\mathrm{reflex}\angle \mathrm{POQ}\right)=120°$