CBSE Class 10 Mathematics Basic 2024: question 29(B)

Hints

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Step-by-step solution

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1. Let $\angle PTQ=\theta$. $TP=TQ$ [tangents from an external point are equal], so $\angle TPQ=\angle TQP$. In $△TPQ$: $\angle TPQ+\angle TQP+\theta =180^{\circ }\Rightarrow \angle TPQ=90^{\circ }-\frac{\theta }{2}$

   1½ marks

2. $OP\perp PT$ [tangent ⟂ radius], so $\angle OPQ+\angle TPQ=90^{\circ }\Rightarrow \angle OPQ=90^{\circ }-\left(90^{\circ }-\frac{\theta }{2}\right)=\frac{\theta }{2}$

   1 mark

3. Hence $\angle PTQ=\theta =2\angle OPQ$. Proved.

   ½ mark

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

Revise the chapter

• Chapter: Circles (notes, formulas, where marks are lost and every past question)