Q) The angle of elevation of a cloud from a point h metres above a lake is α and the angle of depression of its reflection in the lake is β, prove that the distance of the cloud from the point of observation is \frac {2 h \sec \alpha}{(\tan \beta - \tan \alpha)}.

Ans:

Step 1: Let’s draw a diagram for the given question:

The angle of elevation of a cloud from a point h metres above a lake is α 10th Board exam

Here, point of observation is A, cloud is at point B and cloud’s reflection is at point C.

Here, let’s consider the height of AF is h and line AD or Line EF is d, height of cloud from water level is p and CE is also p (reflection).

We need to find length of line AB.

Step 2: Let ‘s start with Δ ADB,

tan ∠ DAB =  \frac{BD}{AD}

∴ tan α = \frac{(p -  h)}{d}

∴ p – h = d tan α

p = h + d tan α  ……….. (i)

Step 3: Next in Δ ADC,

tan ∠ DAC =  \frac{CD}{AD}

∴ tan β = \frac{(p +  h)}{d}

∴ p + h = d tan β

∴ p = d tan β – h ….. (ii)

Step 4: From equations (i) and (ii), we compare values of p:

h + d tan α  = d tan β – h

∴ 2 h = d tan β – d tan α

∴ 2 h = d (tan β – tan α)

∴ d = \frac {2 h}{(\tan \beta - \tan \alpha)} ……… (iii)

Step 5: Δ ADB, cos ∠ DAB =  \frac{AD}{AB}

∴ cos α = \frac{d}{AB}

∴ AB = d sec α …… (iv)

Step 6: By substituting value of d from equation (iii),  in equation (iv), we get:

AB = d sec α

∴ AB = \frac {2 h}{(\tan \beta - \tan \alpha)} \times \sec \alpha

∴ AB = \frac {2 h \sec \alpha}{(\tan \beta - \tan \alpha)}

Hence Proved !

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