Question

If the angle of elevation of a cloud from a point h meters above a lake is a and the angle of depression of its reflection in the lake be b, prove that the distance of the cloud from the point of observation is `(2h sec alpha)/(tan beta - tan alpha)`

Answer 1

Let C′ be the image of cloud C. We have ∠CAB = α and ∠BAC' = β

Again let BC = x and AC be the distance of cloud from point of observation.

We have to prove that

`AC= (2h sec alpha)/(tan beta - tan alpha)`

The corresponding figure is as follows

We use trigonometric ratios.

in Δ ABC

`=> tan alpha = (BC)/(AB)`

`=> tan alpha = x/(AB)`

Again in Δ ABC'

`=> tan beta = (BC')/(AB)`

`=> tan beta = (x + 2h)/(AB)`

Now

`=> tan beta - tan alpha = (x + 2h)/(AB) - x/(AB)`

`=> tan bea - tan alpha = (2h)/(AB)`

`=> AB = (2h)/(tan beta - tan alpha)`

Again in Δ ABC

`=> cos alpha = (AB)/(AC)`

`=> AC = (AB)/(cos alpha)`

`=> (2h sec alpha)/(tan beta - tan alpha)`

Hence distance of cloud from points of observation is `(2h sec alpha)/(tan beta - tan alpha)`