Question

The angle of elevation of a cloud from a point h metres above the level of water in a lake is α and the angle of depression of its reflection in the lake is β. Show that the height of the cloud above the water level is \(\frac{h\,\text{sin}\,(\beta+\alpha)}{\text{sin}\,(\beta-\alpha)}\).

Answer 1

Let CD be the water level. 

Let the height of the cloud above the water level be \(x\) metres, i.e., PD = \(x\) metres. 

As R is the reflection of cloud at P, therefore DR = PD = \(x\) metres. 

Let A be the point h metres above the water level, such that, AC = BD = h metres. 

∴ BP = PD – BD = (\(x\) – h) metres BR = DR + BD = (\(x\) + h) metres. 

Given, ∠PAB = α and ∠BAR = β. 

Let AB = CD = y metres. 

∴ In ∆PAB, tan α = \(\frac{PB}{AB}\) = \(\frac{x-h}{y}\) ⇒ x - h = y tan α         ....(i)

In ∆BAR, tan β = \(\frac{BR}{AB}\) = \(\frac{x+h}{y}\) ⇒ x + h = y tan β          ....(ii)

∴ \(\frac{x+h}{x-h}\) = \(\frac{y\,\text{tan}\,\beta}{y\,\text{tan}\,\alpha}\)         (On dividing eqn. (ii) by eqn. (i))

⇒ \(\frac{2x}{2h}\) = \(\frac{y\,(\text{tan}\,\beta+\text{tan}\,\alpha)}{y\,(\text{tan}\,\beta-\text{tan}\,\alpha)}\)       (Applying componendo and dividendo)

⇒ \(\frac{x}{h}\) = \(\frac{\frac{\text{sin}\,\beta}{\text{cos}\,\beta}+\frac{\text{sin}\,\alpha}{\text{cos}\,\alpha}}{\frac{\text{sin}\,\beta}{\text{cos}\,\beta}-\frac{\text{sin}\,\alpha}{\text{cos}\,\alpha}}\) = \(\frac{\text{sin}\,\beta\,\text{cos}\,\alpha+\text{sin}\,\alpha\,\text{cos}\,\beta}{\text{sin}\,\beta\,\text{cos}\,\alpha-\text{sin}\,\alpha\,\text{cos}\,\beta}\) = \(\frac{\text{sin}\,(\beta+\alpha)}{\text{sin}\,(\beta-\alpha)}\)

⇒ \(x\) = \(\frac{h\,\text{sin}\,(\beta+\alpha)}{\text{sin}\,(\beta-\alpha)}\).