CBSE Class 10 Mathematics Basic 2024: question 33

Hints

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Step-by-step solution

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1. Figure: AB is the tower of height $h$ m, with foot A and top B. C is the point on the opposite bank, with width of canal $CA=x$ m and $\angle BCA=60^{\circ }$. D is 20 m beyond C on line AC, so $DA=x+20$ and $\angle BDA=30^{\circ }$. Angle A $=90^{\circ }$.

   1 mark

2. In right $△BAC$: $\tan 60^{\circ }=\frac{AB}{CA}=\frac{h}{x}$

   1 mark

3. $\sqrt{3}=\frac{h}{x}\Rightarrow h=\sqrt{3}x$ ...(i)

   ½ mark

4. In right $△BAD$: $\tan 30^{\circ }=\frac{h}{x+20}=\frac{1}{\sqrt{3}}$, so $\sqrt{3}h=x+20$ ...(ii)

   1 mark

5. Substituting (i) in (ii): $\sqrt{3}(\sqrt{3}x)=x+20\Rightarrow 3x=x+20\Rightarrow x=10$

   1 mark

6. $h=\sqrt{3}\times 10=1.732\times 10=17.32$ m. Height of the tower = 17.32 m and width of the canal = 10 m.

   ½ mark

Answer: Height of tower = 17.32 m; width of canal = 10 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