CBSE Class 10 Mathematics Basic 2024: question 38

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Step-by-step solution

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1. (i) $AP=AQ$ [tangents from an external point are equal], so $\angle APQ=\angle AQP=\frac{180^{\circ }-60^{\circ }}{2}=60^{\circ }$. Hence $△APQ$ is equilateral.

   ½ mark

2. (i) Therefore $PQ=AP=30$ cm.

   ½ mark

3. (ii) $\angle OPA=\angle OQA=90^{\circ }$ [tangent ⟂ radius]. In quadrilateral APOQ, $\angle POQ=360^{\circ }-90^{\circ }-90^{\circ }-60^{\circ }=120^{\circ }$.

   1 mark

4. (iii)(a) OA bisects $\angle PAQ$, so $\angle PAO=30^{\circ }$.

   ½ mark

5. (iii)(a) In right $△OPA$: $\cos 30^{\circ }=\frac{AP}{OA}\Rightarrow \frac{\sqrt{3}}{2}=\frac{30}{OA}$.

   1 mark

6. (iii)(a) $OA=\frac{60}{\sqrt{3}}=20\sqrt{3}$ cm.

   ½ mark

7. (iii)(b) $\angle PAO=30^{\circ }$ [OA bisects ∠PAQ].

   ½ mark

8. (iii)(b) In right $△OPA$: $\tan 30^{\circ }=\frac{OP}{AP}\Rightarrow \frac{1}{\sqrt{3}}=\frac{OP}{30}$.

   1 mark

9. (iii)(b) Radius $OP=\frac{30}{\sqrt{3}}=10\sqrt{3}$ cm.

   ½ mark

Answer: (i) PQ = 30 cm; (ii) ∠POQ = 120°; (iii)(a) OA = 20√3 cm; (iii)(b) radius = 10√3 cm

Revise the chapter

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