Question

At a point A, 20 metres above the level of water in a lake, the angle of elevation of a cloud is 30 degree. The angle of depression of the reflection of the cloud in the lake, at A is 60 degree. Find the distance of the cloud from A.

Answer 1

in the attachment provided i have solved the answer

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Answer 2

Answer:

The distance of the cloud from A is 40 m.

Step-by-step explanation:

Refer the attached figure

AB = 20 m

AB = ED = 20 m

Let FE =h

FD = EF+ED=20+h

Since BD is the sea level

So, The distance of the object from the sea level is equal to the distance of the image of the object from the sea level

FD = DC =20+h

In Δ AFE

$tan \theta = \frac{Perpendicular}{Base}$

$tan 30^{\circ} = \frac{FE}{AE}$

$\frac{1}{\sqrt{3}}= \frac{h}{AE}$

$AE=\sqrt{3}h$  -1

In Δ AEC

$tan \theta = \frac{Perpendicular}{Base}$

$tan 60^{\circ} = \frac{EC}{AE}$

$tan 60^{\circ} = \frac{ED+DC}{AE}$

$tan 60^{\circ} = \frac{20+20+h}{AE}$

$tan 60^{\circ} = \frac{40+h}{AE}$

$\sqrt{3}= \frac{40+h}{AE}$

$AE= \frac{40+h}{\sqrt{3}}$ --2

Equating 1 and 2

$\sqrt{3}h= \frac{40+h}{\sqrt{3}}$

$3h= 40+h$

$2h= 40$

$20=h$

So, FE = 20 m

In Δ AFE

$Sin \theta = \frac{Perpendicular}{Hypotenuse}$

$Sin 30^{\circ} = \frac{20}{AF}$

$\frac{1}{2}= \frac{20}{AF}$

$AF=40 m$

Hence The distance of the cloud from A is 40 m.