CBSE Class 10 Mathematics Standard 2026: question 31(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. Figure: Draw a circle with centre O and an external point P. Draw tangents PA and PB touching the circle at A and B. Join OA and AB.

   ½ mark

2. $OA\perp AP$ [tangent ⟂ radius at the point of contact], so $\angle OAP=90^{\circ }$.
   Therefore $\angle PAB=90^{\circ }-\angle OAB\text{...(i)}$

   ½ mark

3. $PA=PB$ [tangents from an external point are equal], so $\angle PAB=\angle PBA$ [angles opposite equal sides].
   In $△PAB$, by the angle sum property:
   $\angle APB=180^{\circ }-2\angle PAB$

   1 mark

4. Using (i): $\angle APB=180^{\circ }-2(90^{\circ }-\angle OAB)$

   ½ mark

5. $\Rightarrow \angle APB=2\angle OAB$. Hence proved.

   ½ mark

Answer: Proved: $\angle APB=180^{\circ }-2(90^{\circ }-\angle OAB)=2\angle OAB$.

Revise the chapter

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