In the given figure, $PQ$ and $PR$ are tangents to the circle with centre $O$ such that $\angle QPR = 50^\circ$. Then $\angle OQR$ is equal to:
(a) $25^\circ$
(b) $30^\circ$
(c) $40^\circ$
(d) $50^\circ$
Step-by-Step Solution
Key Concept: $\angle QOR = 180^\circ - 50^\circ = 130^\circ$. In isosceles $\Delta OQR$ ($OQ=OR$), $\angle OQR = \dfrac{180^\circ - 130^\circ}{2} = 25^\circ$.
$\angle QOR = 130^\circ \Rightarrow \angle OQR = (180^\circ - 130^\circ)/2 = 25^\circ$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Evaluating $\angle OQR = 25^\circ$: 1.0 Mark
Correct Answer: $25^\circ$