if the angle of elevation of a cloud from a point ' h ' metres above a lake is alpha and the angle of depression of its reflection in the lake is beeta. prove that the distance of the cloud from the point of observation is      2h sec alpha

                                                               tan beeta - tan alpha??

 
 

Let a be a point h metres above the pond AF and B be the position of the Sky.
We draw a line parallel to EF from A on BD at C.

Clearly, BF = DF

Let, BC = m

so, BF = (m + h)

=> BF = DF = (m + h) metres

Now, in triangle BAC we have,

AB = m cosec α ---- (i)

and, AC = m cot α

Again,

in triangle ACD we have,
AC = (2h + m) cot β

so, m cot α = (2h + m) cot β

=> m = 2h cot β / (cot α -  cot β)

putting, the value of m in (i) we get,

AB = cosec α [2h cot β / (cot α -  cot β)] = 2h sec α / (tan β - tan α) proved..!!


  • 17

what we have to prove plz write properly....

  • 3

T hmbs up again !! 

  • 0

Thank you..!!

  • 6

thank you so much....

  • 1
What are you looking for?