CBSE Class 10 Mathematics Standard 2023: question 32(A)

Hints

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Step-by-step solution

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1. TP = TQ [tangents from an external point are equal], so $\angle TPQ=\angle TQP$ [angles opposite equal sides].

   1 mark

2. Let $\angle PTQ=\theta$. In $△TPQ$, $\angle TPQ+\angle TQP=180^{\circ }-\theta$, so $\angle TPQ=\angle TQP=\frac{180^{\circ }-\theta }{2}=90^{\circ }-\frac{\theta }{2}$.

   1½ marks

3. $\angle OPT=90^{\circ }$ [tangent ⟂ radius], so $\angle OPQ=\angle OPT-\angle TPQ=90^{\circ }-\left(90^{\circ }-\frac{\theta }{2}\right)=\frac{\theta }{2}$.

   1½ marks

4. Therefore $\angle PTQ=\theta =2\angle OPQ$. Hence proved.

   1 mark

Answer: $\angle OPQ=\frac{\theta }{2}$ where $\theta =\angle PTQ$, so $\angle PTQ=2\angle OPQ$.

Revise the chapter

• Chapter: Circles (notes, formulas, where marks are lost and every past question)
• Topic: Solving Problems Using Tangent Properties