CBSE Class 10 Mathematics Basic 2025: question 31

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. Join OP, OQ, OR, OS, where O is the centre. In $\Delta OAP$ and $\Delta OAS$: $OP=OS$ [radii], $OA=OA$ [common], $\angle OPA=\angle OSA=90^{\circ }$ [tangent ⟂ radius]. So $\Delta OAP\cong \Delta OAS$ [RHS], and hence $\angle AOP=\angle AOS$ [CPCT].

   1 mark

2. Similarly, $\angle BOP=\angle BOQ$, $\angle COQ=\angle COR$ and $\angle DOR=\angle DOS$.

   1 mark

3. The angles round O add up to $360^{\circ }$:
   $\angle AOP+\angle BOP+\angle BOQ+\angle COQ+\angle COR+\angle DOR+\angle DOS+\angle AOS=360^{\circ }$
   $\Rightarrow 2(\angle AOP+\angle BOP+\angle COR+\angle DOR)=360^{\circ }$

   ½ mark

4. $\Rightarrow (\angle AOP+\angle BOP)+(\angle COR+\angle DOR)=180^{\circ }$
   $\Rightarrow \angle AOB+\angle COD=180^{\circ }$. Hence proved.

   ½ mark

Answer: ∠AOB + ∠COD = 180° (proved)

Revise the chapter

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