A circle has points P, Q, R on a circle in such a way that angle PQR is $$60^\circ$$ and angle QRP is $$80^\circ$$. Calculate the angle subtended by an arc QR at the centre.

Given : $$\angle$$ PQR = $$60^\circ$$ and $$\angle$$ QRP = $$80^\circ$$

To find : $$\angle$$ QOR = $$\theta$$ = ?

Solution : In $$\triangle$$ PQR, using angle sum property

=> $$\angle$$ PQR + $$\angle$$ QRP + $$\angle$$ QPR = $$180^\circ$$

=> $$60^\circ+80^\circ+$$ $$\angle$$ QPR = $$180^\circ$$

=> $$\angle$$ QPR = $$180^\circ-140^\circ=40^\circ$$

Now, angle subtended by an arc at the centre is double the angle subtended by it at any point on the circle.

=> $$\theta=2\times40^\circ=80^\circ$$

=> Ans - (B)