Gadisar Lake is located in the Jaisalmer district of Rajasthan. It was built by the King of Jaisalmer and rebuilt by Gadsi Singh in 14th century. The lake has many Chhatris. One of them is shown below. From a point A, $h$ m above from water level, the angle of elevation of top of Chhatri (point B) is $45^\circ$ and angle of depression of its reflection in water (point C) is $60^\circ$. If the height of Chhatri above water level is (approximately) $10$ m.
Generated by claude-sonnet-4-6 · 2026-06-15 10:36 · grounding stimulus
Model Answer
(a) Well-labelled figure:
```
B (Top of Chhatri)
|
|
| (10–h)
|
A ————————— D
| \ 45°↑
h | \
| 60°↓ \
Water ————————————
|
C (Reflection of B)
```
- A is at height $h$ above water level.
- B is the top of Chhatri, height = 10 m above water.
- C is the reflection of B, depth = 10 m below water.
- AD is the horizontal distance. Angle of elevation ∠BAD = 45°, angle of depression ∠CAD = 60°.
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(b) Finding h:
Let AD = horizontal distance = $d$.
From the figure:
- $BD = (10 - h)$ (B is above A by this amount)
- $CD = (10 + h)$ (C is the reflection, so 10 m below water, total vertical distance from A = $10 + h$)
From angle of elevation (∠BAD = 45°):
$$\tan 45° = \frac{10 - h}{d} \Rightarrow d = (10 - h) \quad \cdots (1)$$
From angle of depression (∠CAD = 60°):
$$\tan 60° = \frac{10 + h}{d} \Rightarrow \sqrt{3}\,d = (10 + h) \quad \cdots (2)$$
Substituting (1) in (2):
$$\sqrt{3}(10 - h) = 10 + h$$
$$10\sqrt{3} - \sqrt{3}\,h = 10 + h$$
$$10\sqrt{3} - 10 = h + \sqrt{3}\,h = h(1 + \sqrt{3})$$
$$h = \frac{10(\sqrt{3}-1)}{\sqrt{3}+1} = \frac{10(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{10(3 - 2\sqrt{3}+1)}{2} = \frac{10(4-2\times1.73)}{2}$$
$$h = \frac{10(4 - 3.46)}{2} = \frac{10 \times 0.54}{2} = \frac{5.4}{2} \approx \boxed{2.7 \text{ m}}$$
The height of point A above water level is approximately 2.7 m.
Source: Case Study — Trigonometry (Heights and Distances), Chapter 9
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Explanation
- (a) The diagram must show: point A at height h, point B (top of Chhatri) at height 10 m, point C (reflection) at depth 10 m below water, horizontal line AD, and both angles clearly marked. Examiners award marks for correct labelling and positioning.
- (b) Key insight: The reflection C is as far below water as B is above, so vertical distance from A to C = (10 + h). Set up two tan equations and solve simultaneously. Rationalise using $(\sqrt{3}-1)^2 = 4 - 2\sqrt{3}$ to get the numerical answer.