CBSE Class 10 Mathematics Standard 2024: question 32

Hints

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Step-by-step solution

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1. Figure: pillars AB and CD, each of height $h$ m, stand at B and D with BD = 100 m. P is a point on BD with PD = $x$ m and BP = $(100-x)$ m; $\angle CPD=60^{\circ }$, $\angle APB=30^{\circ }$, and $\angle B=\angle D=90^{\circ }$.

   1 mark

2. In right $△CDP$: $\tan 60^{\circ }=\frac{CD}{PD}\Rightarrow \sqrt{3}=\frac{h}{x}$

   1 mark

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

   ½ mark

4. In right $△ABP$: $\tan 30^{\circ }=\frac{AB}{BP}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{100-x}$

   1 mark

5. $\Rightarrow h=\frac{100-x}{\sqrt{3}}$ ...(ii)

   ½ mark

6. From (i) and (ii): $\sqrt{3}x=\frac{100-x}{\sqrt{3}}\Rightarrow 3x=100-x\Rightarrow x=25$

   ½ mark

7. $h=25\sqrt{3}=25\times 1.732=43.3$ m. So each pillar is 43.3 m high, and the point is 25 m from the pillar CD (elevation 60°) and 75 m from the pillar AB (elevation 30°).

   ½ mark

Answer: Height of each pillar = 43.3 m; the point is 25 m and 75 m from the pillars

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