Question

The angle of elevation of a cloud from a point 30m above a lake is 30° and the angle of depression of the reflection of cloud in the lake is 60° , find the height of cloud
and also find the distance of cloud from the point.​

Answer 1

Let AO=H

CD=OB=60m

A'B=AB=(60+H)m

In triangke AOD,

tan30'=AO/OD=H/OD

H=OD/√3

OD=√3 H

Now, in triangle A'OD,

tan60'=OA'/OD=(OB+BA')/OD

√3=(60+60+H)/√3 H

   =(120+H)/√3 H

 =>120+H=3H

     2H=120

      H=60m

Thus, height of the cloud above the lake = AB+A'B

                                                                =(60+60)

                                                                = 120m

Answer 2

Let DE be the level of the lake and A be the point 30m above D (point of the lake)

Let C be the position of the cloud, and F in the reflection of C on water.

Clearly, CE = EF

Using figure, in ∆ABC

tan 30° = $\frac{BC}{AB}$

=> $\frac{1}{\sqrt{3}}$ = $\frac{BC}{AB}$

=> AB = $\sqrt{3}BC$ ----> (1)

Also in ∆ABF

tan 60° = $\frac{BF}{AB}$

=> $\sqrt{3}$ = $\frac{BE + EF}{AB}$

=> $\sqrt{3}$ = $\frac{30 + EF }{AB}$

=> $\sqrt{3}AB$ = 30 + CE {since, CE = EF}

=> $\sqrt{3}AB$ = 30 + BE + BC

=> $\sqrt{3}AB$ = 30 + 30 + BC

=> $\sqrt{3}AB$ = 60 + BC

=> $\sqrt{3} × \sqrt{3}BC$ = 60 + BC {using (1) }

=> 3BC = 60 + BC

=> 2BC = 60

=> BC = 30m

Therefore, CE = BC + BE

=> CE = 30m + 30m = 60m

Hence height of the cloud above the lake is 60m