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Q.

The angle of elevation of a cloud from a point h metres above a lake is $θ$ . The angle of depression of its reflection in the lake is $45^{o}$ . The height of the cloud is

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a

htan (45^{o} + θ)

m

b

2 hcot (45^{o} - θ)

m

c

hcot (45^{o} - θ)

m

d

hcot (45^{o} + θ)

m

answer is A.

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Detailed Solution

Given that,
Height of the point = h m
Angle of elevation =

θ

Angle of depression =

45 °

Question Image

We have the formula,

tanθ = \frac{Height}{Distance}

We have to find the height of the cloud,
Let D be the point above ground CB at a height h. Let A represent the position of the cloud and A’ represent the position of its reflection in water.
Drawing a horizontal line OD parallel to BC. We get OB=CD.
Therefore, the height of the cloud that needs to be found will be AB.

∠ ADO = θ

We have

AO = p

units

∴

In

△ AOD

,

tanθ = \frac{p}{OD}

\Rightarrow OD = \frac{p}{tanθ}

A^{'} B = AB = p + h

In

△ A^{'} OD

\tan 45^{\circ} = \frac{A^{'} O}{OD}

\Rightarrow \tan 45^{\circ} = \frac{A^{'} B + OB}{OD}

\Rightarrow \tan 45^{\circ} = \frac{p + h + h}{OD}

\Rightarrow 1 = \frac{p + 2 h}{p \times cotθ}

\Rightarrow p \times (cotθ - 1) = 2 h

\Rightarrow - p \times (tanθ - 1) = 2 h \times tanθ

\Rightarrow p \times (1 - tanθ) = 2 h \times tanθ

\Rightarrow p = \frac{2 h \times tanθ}{(1 - tanθ)}

Therefore, height of the cloud above the lake

= \frac{2 h \times tanθ}{(1 - tanθ)} + h

Height = (\frac{1 + tanθ}{1 - tanθ}) \times h

\Rightarrow Height = h \times \tan (θ + 45^{\circ})

Thus, the height of the cloud above the lake is

h \times \tan (θ + 45^{\circ}) m

.
Therefore the correct option is 1.

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