CBSE Class 10 Mathematics Standard 2023: question 32

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1. Figure: Let B be the basket and A the point on the ground directly below it, so $AB=h$ m and $\angle BAC=90^{\circ }$. The first observer is at D with $\angle BDA=60^{\circ }$, $AD=x$ m; the second observer is at C on the same side, 100 m beyond D, with $\angle BCA=30^{\circ }$, so $AC=x+100$ m. Let $BD=y$ m.

   1 mark

2. In right $△BAD$: $\tan 60^{\circ }=\frac{h}{x}\Rightarrow \sqrt{3}=\frac{h}{x}\Rightarrow h=\sqrt{3}x$ ……(i)

   1 mark

3. In right $△BAC$: $\tan 30^{\circ }=\frac{h}{x+100}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+100}\Rightarrow x=\sqrt{3}h-100$ ……(ii)

   1 mark

4. (a) Substituting (ii) in (i): $h=\sqrt{3}(\sqrt{3}h-100)=3h-100\sqrt{3}\Rightarrow 2h=100\sqrt{3}\Rightarrow h=50\sqrt{3}$ m ($\approx 86.6$ m).

   ½ mark

5. (b) In $△BAD$: $\sin 60^{\circ }=\frac{h}{y}\Rightarrow \frac{\sqrt{3}}{2}=\frac{50\sqrt{3}}{y}\Rightarrow y=100$ m. So the basket is 100 m from the first observer's eye.

   1 mark

6. (c) $x=\frac{h}{\sqrt{3}}=\frac{50\sqrt{3}}{\sqrt{3}}=50$ m, so the horizontal distance of the second observer from the basket $=AC=x+100=150$ m.

   ½ mark

7. Note: if the two observers are taken on opposite sides of the balloon (DA = x, AC = 100 − x), the same method gives $h=25\sqrt{3}$ m, BD = 50 m and AC = 75 m; the scheme accepts this version as well.

Answer: (a) 50√3 m ≈ 86.6 m; (b) 100 m; (c) 150 m. (Observers on opposite sides: 25√3 m, 50 m, 75 m.)

Revise the chapter

• Chapter: Some Applications of Trigonometry (notes, formulas, where marks are lost and every past question)
• Topic: Line of Sight and Angle of Elevation