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

Given, PQ is a chord of a circle with center O. Also, ∠QPT=60°. Let x be the point on the tangent PT. ∠QPT + ∠OPT = 90° ⇒ ∠OPT = 90° - ∠QPT = 90° - 60° = 30°
In ΔPOQ, ∠POQ = 180° - (∠OPQ + ∠PQO) = 180° - 30° - 30° = 120°
Minor arc ∠POQ = 120°
Therefore Major arc ∠POQ = 360° - 120° = 240°
Angle subtended by an arc at the center is double the angle subtended by it on the remaining part of the circle
∴ ∠QRP = 1/2∠POQ = 120°