CBSE Class 10 Mathematics Standard 2025: question 32(a)

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Step-by-step solution

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1. Figure: Let AB be the lighthouse, with top A and foot B. Ships P and Q are on opposite sides of B on the sea level line PBQ, and $\angle ABP=\angle ABQ=90^{\circ }$.

   The angles of depression from A are $60^{\circ }$ towards P and $45^{\circ }$ towards Q. By alternate angles, $\angle APB=60^{\circ }$ and $\angle AQB=45^{\circ }$.

   $PQ=100\left(\frac{1+\sqrt{3}}{\sqrt{3}}\right)$ m.

   1 mark

2. In right $△ABP$:
   $$\frac{AB}{PB}=\tan 60^{\circ }=\sqrt{3}$$

   1 mark

3. $\Rightarrow PB=\frac{AB}{\sqrt{3}}$ ... (1)

   ½ mark

4. In right $△ABQ$:
   $$\frac{AB}{BQ}=\tan 45^{\circ }=1$$

   ½ mark

5. $\Rightarrow BQ=AB$ ... (2)

   ½ mark

6. Adding (1) and (2):
   $$PQ=PB+BQ=\frac{AB}{\sqrt{3}}+AB=AB\left(\frac{1+\sqrt{3}}{\sqrt{3}}\right)$$

   ½ mark

7. Using the given distance:
   $$100\left(\frac{1+\sqrt{3}}{\sqrt{3}}\right)=AB\left(\frac{1+\sqrt{3}}{\sqrt{3}}\right)$$

   ½ mark

8. $\Rightarrow AB=100$ m.

   So the height of the lighthouse is 100 m.

   ½ mark

Answer: Height of the lighthouse = 100 m

Revise the chapter

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