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$\sf{Question}$

If the angle of the elevation of a cloud from a point $\sf{h}$ above a lake is α and the angle of depression of its reflection in the lake is β , prove that height of the cloud is

$\sf{\frac{h (tan β + tan α)}{tan β - tan α}$

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Answer 1

Let AB be the surface of the lake and let P be a point of observation such that AP = h metres.

Let C be the position of the cloud and C' be its reflection in the lake.

Then, CB=C'B. Let PM be perpendicular from P on CB. Then, ∠CPM= α and ∠MPC' = β. Let CM = x.

Then, CB=CM+MB=CM+PA = x+h

In △ CPM, we have,

tanα=

PM

CM

tanα=

AB

x

AB=xcotα .....(1)

In △PMC', we have

tanβ=

PM

C

′

M

tanβ=

AB

x+2h

AB=(x+2h)cotβ .....(2)

From (1) & (2), we have,

xcotα=(x+2h)cotβ

x(cotα−cotβ)=2hcotβ

x(

tanα

1

−

tanβ

1

)=

tanβ

2h

x(

tanαtanβ

tanβ−tanα

)=

tanβ

2h

x=

tanβ−tanα

2htanα

Hence, the height CB of the cloud is given by

CB=x+h

CB=

tanβ−tanα

2htanα

+h

CB=

tanβ−tanα

2htanα+htanβ−htanα

=

tanβ−tanα

h(tanα+tanβ)

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Answer 2

Let AB be the surface of the lake and let P be a point of observation such that AP = h metres.

Let C be the position of the cloud and C' be its reflection in the lake.

Then, CB=C'B. Let PM be perpendicular from P on CB. Then, ∠CPM= α and ∠MPC' = β. Let CM = x.

Then, CB=CM+MB=CM+PA = x+h

In △ CPM, we have,

tanα=

PM

CM

tanα=

AB

x

AB=xcotα .....(1)

In △PMC', we have

tanβ=

PM

C

′

M

tanβ=

AB

x+2h

AB=(x+2h)cotβ .....(2)

From (1) & (2), we have,

xcotα=(x+2h)cotβ

x(cotα−cotβ)=2hcotβ

x(

tanα

1

−

tanβ

1

)=

tanβ

2h

x(

tanαtanβ

tanβ−tanα

)=

tanβ

2h

x=

tanβ−tanα

2htanα

Hence, the height CB of the cloud is given by

CB=x+h

CB=

tanβ−tanα

2htanα

+h

CB=

tanβ−tanα

2htanα+htanβ−htanα

=

tanβ−tanα

h(tanα+tanβ)