Q1. [4] § 9.1 Heights and Distances § 9.2 Summary
Amrita stood near the base of a lighthouse, gazing up at its towering height. She measured the angle of elevation to the top and found it to be 60°. Then, she climbed a nearby observation deck, 40 metres higher than her original position and noticed the angle of elevation to the top of lighthouse to be 45°.
Based on the above given information, answer the following questions:
- (i) Find $CD$ in terms of $h$ (where $h$ is the height). [1]
- (ii) Find $BD$ in terms of $BC$. [1]
- (iii) Find the height $CE$ of the lighthouse. [Use $\sqrt{3} = 1{\cdot}73$] [2]
Previously asked in CBSE board exam
2025 30/1/1 Q38
Generated by claude-sonnet-4-6 · 2026-06-15 10:34 · grounding stimulus
Model Answer
Let CE = height of lighthouse, CD = horizontal distance, DE = h (height above D), so CE = CD + DE...
Setup: Let BC = h (height of lighthouse above observation deck level), so BE = h. Let CD = horizontal distance from Amrita to lighthouse base.
(i) From point B (observation deck), angle of elevation = 45°:
$$\tan 45° = \frac{BE}{BD} \Rightarrow 1 = \frac{h}{BD} \Rightarrow BD = h$$
$$\therefore CD = BD = h$$
(ii) From point B, angle of elevation = 45°:
$$\tan 45° = \frac{BE}{BD} = 1 \Rightarrow BD = BE = BC$$
$$\therefore BD = BC$$
(iii) From point A (ground level), angle of elevation = 60°. AB = 40 m, so AE = h + 40:
$$\tan 60° = \frac{h + 40}{CD} \Rightarrow \sqrt{3} = \frac{h + 40}{h}$$
$$\sqrt{3}\,h = h + 40 \Rightarrow h(\sqrt{3} - 1) = 40$$
$$h = \frac{40}{\sqrt{3}-1} = \frac{40(\sqrt{3}+1)}{2} = 20(\sqrt{3}+1)$$
$$h = 20(1.73 + 1) = 20 \times 2.73 = 54.6 \text{ m}$$
$$\therefore CE = h + 40 = 54.6 + 40 = \textbf{94.6 m}$$
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Explanation
- The key is setting BC = h (lighthouse height above deck level B) and noting CD = BD since tan 45° = 1.
- For part (iii), from A the full height seen is (h + 40) over horizontal distance h, giving the equation with √3.
- Rationalize the denominator properly. Examiners check the rationalization step and correct substitution of √3 = 1.73.
- Label all parts clearly — examiners award step marks for correct tan ratio setup, equation, and final answer.
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cbse.gov.in.