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CBSE Class 10 Basic Math 2025 All Sets Solved Paper

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Question : 11 of 20
Marks: +1, -0
PQ and PR are tangents to a circle with centre O such that OQ = QP. The value of ∠OPQ\angle OPQ is equal to
Solution:  
Tangents from an external point are equal.
PQ = PR
The radius of the circle at the point of tangency is perpendicular to the tangent.
∠OQP=90∘\angle O Q P = 90^{\circ}
Since, OQ = QP
So, â–³ O Q P is isosceles.
∠POQ=∠OPQ\angle P O Q = \angle O P Q
Inâ–³OQP\text{In} \triangle O Q P
∠OQP+∠POQ+∠OPQ=180∘\angle O Q P + \angle P O Q + \angle O P Q = 180^{\circ}
90∘+2∠OPQ=180∘90^{\circ} + 2 \angle O P Q = 180^{\circ}
2∠OPQ=180∘−90∘2 \angle O P Q = 180^{\circ} - 90^{\circ}
∠OPQ=90∘2\angle O P Q = \frac{90^{\circ}}{2}
∴∠OPQ=45∘\therefore \angle O P Q = 45^{\circ}
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