Question

If the angle of elevation of a cloud from a point h m above a lake is α and the angle of depression of its reflection in the lake be β. Prove that the distance of the cloud from the point of observations is 2hsecα/(tanβ - tanα).

Answer 1

Let LC is a surface of lake and D is a point of observation. A is the position of cloud.

The angle of elevation from point D is α.

Let AC = H

Again, let D = y, and DB = x. 

The angle of depression of shadow of cloud from point D is β.

Hence, ∠ADB = α and ∠BDA’ = β

From right angled ∆ABD,

tan α = AB/DB

⇒ tan α = (H - h)/x

⇒ H – h = x tan α

⇒ H = h + x tan α ….(i)

From right angled ∆A’BD,

tan β = A'B/DB

⇒ tan β = (H - h)/x

⇒ H + h = x tan β

Put the value of H from equation (i) in equation (ii),

h + x tan α + h = x tan β

⇒ x tan α + 2h = x tan β

⇒ 2h = x tan β – x tan α

⇒ 2h = x(tan β – tan α)

⇒ x = 2h/(tan β - tanα) ….(iii)

From right angled ∆ABD,

cos α = DB/DA

= x/y

y = cotθ = x/cosα

= x sec α ……(iv)

Put the value of x from equation (iii) in equation (iv)

y = 2hsecα/(tanβ - tanα)

Hence, distance of cloud from the point of observation is 2hsecα/(tanβ - tanα).