CBSE Class 10 Mathematics Standard 2025: question 26(b)

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, and join OA, OB and AB.

   To prove: $\angle APB+\angle AOB=180^{\circ }$.

   1 mark

2. $OA\perp PA$ and $OB\perp PB$ [tangent ⟂ radius at the point of contact].

   So $\angle OAP=\angle OBP=90^{\circ }$.

   1 mark

3. In quadrilateral OAPB, the angles add up to $360^{\circ }$:
   $$\angle APB+\angle OAP+\angle OBP+\angle AOB=360^{\circ }$$
   $$\angle APB+90^{\circ }+90^{\circ }+\angle AOB=360^{\circ }$$

   ½ mark

4. So $\angle APB+\angle AOB=180^{\circ }$.

   Hence $\angle APB$ and $\angle AOB$ are supplementary.

   ½ mark

Answer: $\angle APB+\angle AOB=180^{\circ }$, so the two angles are supplementary

Revise the chapter

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