CBSE Class 10 • 2024 • 5 Marks

Some Applications of Trigonometry: Angle of Elevation and Depression of Cloud & Lake Reflection

Official examination question with verified M1/A1 mark scheme and step-by-step mathematical reasoning.

Problem Statement

The angle of elevation of a cloud from a point $h$ meters above the surface of a lake is $\alpha$ and the angle of depression of its reflection in the lake is $\beta$. Prove that the height of the cloud above the surface of the lake is: $h \left( \frac{\tan \beta + \tan \alpha}{\tan \beta - \tan \alpha} \right)$

Verified Solution & Marking Scheme

Set up Geometric Diagram Relations
Let the surface of the lake be the horizontal reference line. - Point of observation $O$ is at height $h$ above the lake surface. - Let the cloud $C$ be at height $H$ above the lake surface. - Height of cloud above observer $O$ is $H - h$. - The reflection $R$ of the cloud in the lake is at depth $H$ below the surface. - Distance of reflection $R$ below the observer $O$ is $H + h$. - Let the horizontal distance from the observer to the vertical line of the cloud be $d$.
Write Trigonometric Ratios for Elevation and Depression
From the right triangle formed with the cloud: $\tan \alpha = \frac{H - h}{d} \implies d = \frac{H - h}{\tan \alpha}$ ...(1) From the right triangle formed with the reflection: $\tan \beta = \frac{H + h}{d} \implies d = \frac{H + h}{\tan \beta}$ ...(2)
Equate Horizontal Distance d and Solve for H
Equating (1) and (2): $\frac{H - h}{\tan \alpha} = \frac{H + h}{\tan \beta}$ $(H - h) \tan \beta = (H + h) \tan \alpha$ $H \tan \beta - h \tan \beta = H \tan \alpha + h \tan \alpha$ $H (\tan \beta - \tan \alpha) = h (\tan \beta + \tan \alpha)$ $H = h \left( \frac{\tan \beta + \tan \alpha}{\tan \beta - \tan \alpha} \right)$.
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