The angle of elevation of a cloud from a point ‘h’ meters above a lake is

α
and the angle of depression of its reflection in the lake is

β
. Then the distance of the cloud from the point of observation is ____.

Question

The angle of elevation of a cloud from a point ‘h’ meters above a lake is

α
  and the angle of depression of its reflection in the lake is

β
  . Then the distance of the cloud from the point of observation is ____.

Accepted Answer

According to the above figure:Let A be the cloud and the observer is at point B which is ‘h’ metres above the lake level. Let the reflection of the cloud below the lake level be at point C. Point D and E are at lake level.We have assumed the height of the cloud above the lake level as ‘x’. Therefore, AE = x. Also, assume that the distance BM is ‘y’.Now, in right angle triangle ABM,∠ABM = αWe know that,

tanθ=
    
     perpendicular
    
     base

. Therefore,

tanα=
    
     AM
    
     BM

.Since, BD = ME = h, because they are opposite sides of the rectangle BDEM. Therefore,AM = AE – EM = x – h.

⇒tanα=
      
       AM
      
       BM

⇒tanα=
      
       x−h
      y

⇒y=
      
       x−h
      
       tanα
     
     .........(1)

We know that,

secθ=
    
     hypotenuse
    
     base

. Therefore,

secα=
    
     AM
    
     BM

.Since, BD = ME = h, because they are opposite sides of the rectangle BDEM. Therefore,AM = AE – EM = x – h.

⇒secα=
      
       AM
      
       BM

⇒tanα=
      d
      y

⇒y=
      d
      
       secα
     
     .........(2)

Substituting the value of y from equation (ii) in equation (i), we get,

x−h
    
     tanα
   
   =
    d
    
     secα
   
   .........(3)
  Now, in right angle triangle BMC,∠CBM = βWe know that,

tanθ=
    
     perpendicular
    
     base

. Therefore,

tanβ=
    
     CM
    
     BM

.We know that, from the laws of angle of reflection we have, the distance of the object and its image formed by the plane mirror are equal. Therefore, considering the lake as the plane mirror, we have,AE = CE = x.CM = CE + ME = h + x

⇒tanβ=
      
       h+x
      y

⇒y=
      
       h+x
      
       tanβ

⇒
      d
      
       secα
     
     =
      
       h+x
      
       tanβ

⇒
      
       dtanβ
      
       secα
     
     =h+x

⇒x=
      
       dtanβ
      
       secα
     
     −h………(4)

Substituting the value of

x
   from equation (4) in equation (3), we get,

d
      
       secα
     
     =

dtanβ
        
         secα
       
       −h−h
      
       tanα

⇒
      d
      
       secα
     
     =

dtanβ
        
         secα
       
       −2h
      
       tanα

By cross-multiplication we get,

dtanα=dtanβ−2hsecα

⇒dtanβ−dtanα=2hsecα

⇒d(tanβ−tanα)=2hsecα

⇒d=
      
       2hsecα
      
       (tanβ−tanα)

The angle of elevation of a cloud from a point ‘h’ meters above a lake is $α$ and the angle of depression of its reflection in the lake is $β$ . Then the distance of the cloud from the point of observation is ____.

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The angle of elevation of a cloud from a point ‘h’ meters above a lake is α and the angle of depression of its reflection in the lake is β . Then the distance of the cloud from the point of observation is ____.

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The angle of elevation of a cloud from a point ‘h’ meters above a lake is $α$ and the angle of depression of its reflection in the lake is $β$ . Then the distance of the cloud from the point of observation is ____.