CBSE Class 10 Mathematics Standard 2023: question 29(a)

Hints

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Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. $TP=TQ$ [tangents drawn from an external point are equal].

   So in $△TPQ$, $\angle TPQ=\angle TQP$ [angles opposite equal sides].

   1 mark

2. Let $\angle PTQ=\theta$. By the angle sum property of $△TPQ$:
   $\angle TPQ=\angle TQP=\frac{180^{\circ }-\theta }{2}=90^{\circ }-\frac{\theta }{2}$.

   1 mark

3. $\angle OPT=90^{\circ }$ [tangent $\perp$ radius at the point of contact].

   So $\angle OPQ=90^{\circ }-\angle TPQ=90^{\circ }-\left(90^{\circ }-\frac{\theta }{2}\right)=\frac{\theta }{2}$.

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

   1 mark

Answer: $\angle PTQ=2\angle OPQ$ (proved)

Revise the chapter

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