Given, O is the centre of the given circle
∴ OQ and OP are the radius of the circle.
∴ PT is a tangent
∴ OP⊥PT
(Radius of a circle is perpendicular to the tangent through the point of contact)
So, ∠OPT=90∘
∠OPQ=90∘−∠QPT
∠OPQ=90∘−60∘
[Given,∠QPT=60∘]
∠OPQ=30∘
∴ ∠OQP=30∘[∵ΔOPQ is isosceles triangle]
Now,in ΔOPQ
∠POQ+∠OPQ+∠OQP=180∘
∠POQ+30∘+30∘=180∘
∠POQ=120∘
reflex ∠POQ=360∘−120∘=240∘
∴ ∠PRQ=12 reflex ∠POQ
[∴ The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.]
∠PRQ=12×240∘
Hence, ∠PRQ=120∘