CBSE Class 10 Mathematics Basic 2023: question 37

Hints

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Step-by-step solution

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1. (i) $OA\perp NA$ [tangent ⟂ radius], so in right $△OAN$: $\tan 30^{\circ }=\frac{OA}{AN}\Rightarrow \frac{1}{\sqrt{3}}=\frac{5}{AN}$

   ½ mark

2. $\Rightarrow AN=5\sqrt{3}$ cm

   ½ mark

3. (ii) NO bisects $\angle ANB$ [tangents from an external point], so $\angle BNO=30^{\circ }$ and $\angle ANB=60^{\circ }$. In quadrilateral OANB, $\angle OAN=\angle OBN=90^{\circ }$, so $\angle AOB=360^{\circ }-90^{\circ }-90^{\circ }-60^{\circ }=120^{\circ }$.

   1 mark

4. (iii)(a) $NA=NB$ [tangents from N] and $\angle ANB=60^{\circ }$, so $\angle NAB=\angle NBA=60^{\circ }$; $△NAB$ is equilateral.

   ½ mark

5. Hence $AB=NA=NB=5\sqrt{3}$ cm.

   1 mark

6. Total length $=NA+NB+AB=3\times 5\sqrt{3}=15\sqrt{3}$ cm $\approx 25.98$ cm.

   ½ mark

7. (iii)(b) If $\angle ANO=45^{\circ }$, then $\angle ANB=90^{\circ }$, and $\angle AOB=360^{\circ }-90^{\circ }-90^{\circ }-90^{\circ }=90^{\circ }$.

   ½ mark

8. So every angle of OANB is $90^{\circ }$ ($\angle OAN=\angle OBN=90^{\circ }$ as tangent ⟂ radius).

   1 mark

9. Also $OA=OB$ (radii) and $NA=NB$; in right $△OAN$ with $45^{\circ }$, $NA=OA=5$ cm. All sides equal and all angles right, so OANB is a square.

   ½ mark

Answer: (i) AN = 5√3 cm; (ii) ∠AOB = 120°; (iii)(a) 15√3 cm (≈ 25.98 cm); (iii)(b) OANB is a square

Revise the chapter

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