question_answer

If the angle of elevation of a cloud from a point h metres above a lake is x and the angle of depression of its reflection in the lake is y, then the distance of the cloud from the point of observation is

A)  \[\frac{2h\,\sec \,x}{\tan y+\tan x}\] 

B)  \[\frac{2h\,cos\,x}{\tan x+\tan \,y}\]

C)                     \[\frac{2h\,cot\,x}{\tan \,y+\tan \,x}\]

D)         \[\frac{2h\,\sec \,x}{\tan \,y-\tan \,x}\]

Correct Answer: D

Solution :

Let AB be the surface of the lake and let C be a point of observation such that AC = h metres. Let D be the position of the cloud and D' be its reflection in the lake. Then BD = B'. In \[\Delta \,DEC,\] \[\tan x=\frac{DE}{CE}\Rightarrow CE=\frac{H}{\tan x}\]   ?(1) In \[\Delta \,CED',\tan y=\frac{ED'}{EC}\Rightarrow CE=\frac{h+H+h}{\tan y}\] \[\Rightarrow \]            \[CE=\frac{2h+H}{\tan y}\] From (1) and (2), \[\frac{H}{\tan x}=\frac{2h+H}{\tan y}\] \[\Rightarrow \]            \[H\tan y=2h\,\tan x+H\tan x\] \[\Rightarrow \]            \[H(\tan y-\tan x)=2h\,\tan x\] \[\Rightarrow \]            \[H=\frac{2h\,\tan \,x}{\tan y-\tan x}\]           ??(3) In \[\Delta \,DCE,\]   \[\sin x=\frac{DE}{CD}\] \[\Rightarrow \] \[CD=\frac{DE}{\sin x}\Rightarrow CD=\frac{H}{\sin x}\] Substituting the value of H from (3), we have \[CD=\frac{2h\,\,\tan x}{(\tan y-\tan x)\sin x}=\frac{2h\,\sec x}{\tan y-\tan x}\] Hence, the distance of the cloud from the point of observation is \[\frac{2h\,\,\sec x}{\tan y-\tan \,x}\]