CBSE Class 10 Mathematics Standard 2026: question 29(b)

Hints

Open one hint at a time and try again before the next.

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. Given: TP and TQ are tangents from T to a circle with centre O. To prove: $\angle PTQ=2\angle OPQ$.

   Let $\angle TPQ=\theta$. Since $\angle OPT=90^{\circ }$ [tangent ⟂ radius at the point of contact], we get $\angle OPQ=90^{\circ }-\theta$ ...(1)

   ½ mark

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

   ½ mark

3. So $\angle TQP=\angle TPQ=\theta$ [angles opposite equal sides of a triangle].

   ½ mark

4. In $△PTQ$: $\angle TPQ+\angle TQP+\angle PTQ=180^{\circ }$ [angle sum property]

   $\Rightarrow \theta +\theta +\angle PTQ=180^{\circ }$

   ½ mark

5. $\Rightarrow \angle PTQ=180^{\circ }-2\theta =2(90^{\circ }-\theta )$

   ½ mark

6. Using (1), $\angle PTQ=2\angle OPQ$. Hence proved.

   ½ mark

Answer: Proved: ∠PTQ = 2∠OPQ

Revise the chapter

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