P and Q are two points on a circle with centre at O. R is a point on the minor arc of the circle, between the points P and Q. The tangents to the circle at the points P and Q meet each other at the point S. If ∠PSQ = 20°, ∠PRQ = ?

It is given that $$\angle$$PSQ = 20°

and we know that $$\angle$$OPS = $$\angle$$OQS = 90°

Now, in quadrilateral OPQS, sum of all angles is 360°

=> $$\angle$$POQ + $$\angle$$PSQ + $$\angle$$OPS + $$\angle$$OQS = 360°

=> $$\angle$$POQ + 20° + 90° + 90° = 360°

=> $$\angle$$POQ = 160°

Using the property that angle subtended at the centre is double the angle subtended by the same points at any other point on the circle.

=> $$\angle$$PTQ = $$\frac{\angle POQ}{2}$$ = 80°

Also, PTQR is cyclic quadrilateral and thus sum of opposite angles is 180°

=> $$\angle$$PRQ = 180° - $$\angle$$PTQ = 180° - 80° = 100°