From an aeroplane vertically above a …

Let us suppose that aeroplane is at A

Suppose B and  C are two consecutive kilometre stones such that
{tex}\angle \mathbf { X } \mathbf { A } \mathbf { B } = \alpha{/tex} and {tex}\angle \mathrm { YAC } = \beta{/tex}
{tex}\therefore \angle A B D = \alpha{/tex} and {tex}\angle A C D = \beta{/tex}
Let us suppose that BD = x km.
AD is height of aeroplane.
In {tex}\triangle{/tex}ADB, {tex}\frac { A D } { B D } = \tan \alpha{/tex}
{tex}\Rightarrow \quad \frac { \mathrm { AD } } { x } = \tan \alpha {/tex}     

{tex}\Rightarrow x = \frac { \mathrm { AD } } { \tan \alpha }{/tex}

In {tex}\triangle{/tex}ADC, {tex}\frac { \mathrm { AD } } { \mathrm { DC } } = \tan \beta{/tex}
{tex}\Rightarrow \quad \frac { \mathrm { AD } } { 1 - x } = \tan \beta{/tex}
{tex}\Rightarrow \frac { \mathrm { AD } } { 1 - \frac { \mathrm { AD } } { \tan \alpha } } = \tan \beta{/tex}
{tex}\Rightarrow \frac { \operatorname { AD } \tan \alpha } { \tan \alpha - \mathrm { AD } } = \tan \beta{/tex}
{tex}\Rightarrow A D \tan \alpha = \tan \alpha . \tan \beta - A D \tan \beta{/tex}
{tex}\Rightarrow A D \tan \alpha + A D \tan \beta = \tan \alpha . \tan \beta{/tex}
{tex}\Rightarrow \mathbf { A D } = \frac { \tan \alpha \cdot \tan \beta } { \tan \alpha + \tan \beta }{/tex}                 
Hence proved.