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Question
If the angle of elevation of a cloud from a point hh metres above a lake is α\alpha and the angle of depression of its reflection in the lake is β\beta, the height of the cloud above the lake is:
(A)hsin⁡(α+β)sin⁡(β−α)\dfrac{h \sin(\alpha + \beta)}{\sin(\beta - \alpha)}
(B)hsin⁡(β−α)sin⁡(α+β)\dfrac{h \sin(\beta - \alpha)}{\sin(\alpha + \beta)}
(C)htan⁡αtan⁡β−tan⁡α\dfrac{h \tan\alpha}{\tan\beta - \tan\alpha}
(D)h(tan⁡α+tan⁡β)tan⁡α−tan⁡β\dfrac{h(\tan\alpha + \tan\beta)}{\tan\alpha - \tan\beta}
Solution Path
Cloud at height HH, reflection at depth HH. Set up tan⁡α=(H−h)/d\tan\alpha = (H-h)/d and tan⁡β=(H+h)/d\tan\beta = (H+h)/d. Componendo-dividendo gives H=hsin⁡(α+β)/sin⁡(β−α)H = h\sin(\alpha+\beta)/\sin(\beta-\alpha).
01Question Setup
1/4
A cloud at height HH above a lake is observed from hh metres above the lake. Elevation to cloud is α\alpha, depression to reflection is β\beta. Find HH.
Reflection appears at depth HH below lake surface
02Diagram with Lake, Cloud, Reflection
2/4
The observer is hh above the lake. The cloud is H−hH - h above the observer. The reflection is H+hH + h below the observer. Both share horizontal distance dd.
tan⁡α=H−hd\tan\alpha = \frac{H-h}{d} and tan⁡β=H+hd\tan\beta = \frac{H+h}{d}
03Two Equations, Componendo-DividendoKEY INSIGHT
3/4
Divide the two tan equations to eliminate dd. Apply componendo-dividendo: tan⁡βtan⁡α=H+hH−h\frac{\tan\beta}{\tan\alpha} = \frac{H+h}{H-h} gives HH in terms of hh, α\alpha, β\beta.
H=h⋅tan⁡β+tan⁡αtan⁡β−tan⁡αH = h \cdot \frac{\tan\beta + \tan\alpha}{\tan\beta - \tan\alpha}
04Final Answer
4/4
Converting to sine form: H=hsin⁡(α+β)sin⁡(β−α)H = \frac{h \sin(\alpha + \beta)}{\sin(\beta - \alpha)}.
Answer (A):H=hsin⁡(α+β)sin⁡(β−α)\boxed{\text{Answer (A)}: H = \frac{h\sin(\alpha+\beta)}{\sin(\beta-\alpha)}}
Concepts from this question2 concepts unlocked
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Cloud and Reflection in a Lake

TRICKY

A cloud at height h above a lake and its reflection at depth h below the surface are observed from a point at height a above the lake. The elevation angle (to the cloud) and depression angle (to the reflection) give two equations that determine h.

h=a⋅tan⁡α+tan⁡βtan⁡β−tan⁡αh = a \cdot \frac{\tan\alpha + \tan\beta}{\tan\beta - \tan\alpha}

This is a classic JEE problem type. The reflection appears at the same depth below the lake as the cloud is above it. Setting up (h - a) for elevation and (h + a) for depression is the key step most students miss.

Cloud height problemsReflection in water surfaceObservation from cliff above lake
Practice (5 Qs) →

Elevation and Depression Triangle Setup

EASY

The angle of elevation is the angle between the horizontal and the line of sight when looking upward. The angle of depression is the angle when looking downward. Both create right triangles where tan(angle) = opposite/adjacent is the primary relation.

tan⁡(α)=heightdistance,Elevation angle=Depression angle (alternate angles)\tan(\alpha) = \frac{\text{height}}{\text{distance}}, \quad \text{Elevation angle} = \text{Depression angle (alternate angles)}

Every heights and distances problem starts with identifying whether you are looking up or down. Getting this wrong means the entire triangle is set up incorrectly. The alternate angle property is the single most used fact in this chapter.

Basic height problemsDepression distance problemsTower height from ground
Practice (12 Qs) →
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