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 The angle of elevation of a cloud from a point 60m above a lake is 30degree and the angle of depression of the reflection of the cloud in the lake is 60degree. plz find the height of the cloud.

Let AO = H

CD = OB = 60 m

A' B = AB = 60 + H

 

In ∆ AOD,

Now, In ∆ A'OD,

⇒ 120 + H = 3H

⇒ 2H = 120

⇒ H = 60 m

 

Thus, height of the cloud above the lake = AB + A'B = 60 + 60 = 120 m.


 

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 is there no one to answer me

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 Let AB be the surface of the lake and Let P be a point vertically above A such that AP = 60m. 

Let C be the position of the cloud and let D be its reflection in the lake. 
Let the height of the cloud be H metres.

Then, BC = BD = H,
Draw PQ perpendicular to CD. Then, angle QPC = 30 degree, angle QPD = 60 degree, BQ = AP = 60m.

CQ = (H-60) m and DQ = (H+60) m. 
From right triangle CQP, we have
PQ/CQ = cot 30 degree 
=> PQ/(H-60)=root 3
=> PQ = (H-60) root3 m ....(i)

From the right triangle DQP, we have
PQ/DQ = cot 60degree 
=>PQ/(H+60)=1/root3
=> PQ = (H+60)/root3 m ....(ii)

From (i) and (ii), we get
(H-60)root3 
= (H+60)/root3
=> (3H-180) = (H+60)
=> 3H - H = 60+180 => 2H = 240
=> H = 120
Hence, the height of the cloud is 120 m. 

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plz draw figure

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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..!!

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This was the solution to some some previous questions posted by someone...

Just get the idea from the figure...11

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here is the answer368_16302_trigo.jpg

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kgh

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 Rahul Mishra i didn't understood the last line of of your solution . could you please solve that list lline briefly? please

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find height of cloud above the ground
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