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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.
=> Ans - (B)
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