Question

The angle of elevation of a cloud from a point h metres above a lake is alpha and the angle of depression of reflection of a cloud is in the lake is beta prove the distance of the cloud from the point of observation is to hatch second alpha divided by 10 beta minus tan alpha metre

Answer 1

Answer:

Let us assume (as shown in the figure below)

The surface of the lake to be “AB”

The position of the observer at h meters above the lake be “P”

The position of the cloud be “C” and its reflection be at “ C’ ”. Then, CB= C’B

The angle of elevation i.e., ∠CPQ = α and angle of depression i.e., ∠C’PQ = β.  

Length CQ be “x” meters

Considering ∆PQC, since ∠Q is 90°, therefore we can write

sin α = perpendicular / hypotenuse = CQ / CP = x / CP

or, x = CP sin α ……. (i)

and,

tan α = perpendicular / base = CQ / PQ = x / PQ

or, PQ = x / tan α …… (ii)

Now considering ∆PQC’, we can write

tan β = QC’ / PQ = [QB + BC’]/PQ = [h+x+h]/PQ  

[∵ from the figure we have PA=QB= h meters, CQ= x meters therefore, BC’ = (x+h) meters]

or, tan β = [2h + x]/PQ

or, PQ = [2h+x] / tan β ……. (iii)

Comparing (ii) & (iii), we get

x / tan α = [2h+x] / tan β

or, x tan β = [2h+x] tan α

or, x tan β = 2h tan α + x tan α

or, x tan β - x tan α = 2h tan α

or, x (tan β - tan α) = 2h tan α

or, x = 2h tan α / (tan β - tan α) …… (iv)

Comparing (i) and (iv), we get

CP sin α = 2h tan α / (tan β - tan α)

Or, CP = [2h (sin α / cos α)] / [sin α (tan β - tan α)]

Or, CP = [2h/cos α] * [sin α / sin α] * [1 / (tan β - tan α)]

Or, CP = [2h sec α] / [tan β - tan α] = distance of the cloud from the point of observer

Hence proved, the distance of the cloud from the point of observation is [2h sec α] / [tan β - tan α].