Elevated water storage tanks are built to store and supply water to nearby colonies. In the diagram given above, $AB$ is an elevated water tank and $CD$ is a nearby multistorey building. The building is 54 metres away from the water tank. From a window ($W$) of the building, the angle of elevation of top of the tank is $45°$ and angle of depression of its foot is $30°$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:38 · grounding stimulus
Model Answer
(i) From window W, angle of depression of foot B is 30°.
$$\tan 30° = \frac{d}{54} \Rightarrow \frac{1}{\sqrt{3}} = \frac{d}{54}$$
So the relation is: $d = \frac{54}{\sqrt{3}}$ i.e., $y = d = \frac{54}{\sqrt{3}} = 18\sqrt{3}$ m
(ii) From window W, angle of elevation of top A is 45°.
$$\tan 45° = \frac{h}{54} \Rightarrow 1 = \frac{h}{54} \Rightarrow h = 54 \text{ m}$$
(iii) Height of water tank AB = $h + d = 54 + 18\sqrt{3}$ m
$$= 18(3 + \sqrt{3}) = 18\sqrt{3}(\sqrt{3}+1) \text{ m}$$
OR
$x = 54$ m (horizontal distance, as $\tan 45° = h/x \Rightarrow x = h = 54$ m)
Height of window above ground $= d = 18\sqrt{3}$ m $\approx 31.18$ m
Source: Heights and Distances, Application of Trigonometry
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Explanation
- The horizontal distance (54 m) is used with both angles to set up tan equations.
- $h$ is the vertical height from window level up to top of tank; $d$ is from ground to window (same as depression side).
- Total tank height = $h + d$ since B is at ground and A is at top.
- Examiners expect clear trigonometric ratios, substitution, and final simplified value.