From an aeroplane vertically above a straight horizontal road,

Let $h$ be the height of aero plane $P$ above the road. And $A$ and $B$ be the two consecutive milestone, then $A B=1$ mile. We have $\angle P A Q=\alpha$ and $\angle P B Q=\beta$.

We have to prove that

$h=\frac{\tan \alpha \tan \beta}{\tan \alpha+\tan \beta}$

The corresponding figure is as follows

In $\triangle P A Q$

$\Rightarrow \quad \tan \alpha=\frac{P Q}{A Q}$

$\Rightarrow \quad \tan \alpha=\frac{h}{x}$

$\Rightarrow \quad x=\frac{h}{\tan \alpha}$

$\Rightarrow \quad x=h \cot \alpha$

Again in $\triangle P B Q$

$\Rightarrow \quad \tan \beta=\frac{P Q}{B Q}$

$\Rightarrow \quad \tan \beta=\frac{h}{y}$

$\Rightarrow \quad y=\frac{h}{h}$

$\Rightarrow \quad y=\frac{h}{\tan \beta}$

$\Rightarrow \quad y=h \cot \beta$

Now,

$\Rightarrow \quad A B=x+y$

$\Rightarrow \quad A B=h(\cot \alpha+\cot \beta)$

$\Rightarrow \quad A B=h\left(\frac{1}{\tan \alpha}+\frac{1}{\tan \beta}\right)$

$\Rightarrow \quad A B=h\left(\frac{\tan \alpha+\tan \beta}{\tan \alpha \tan \beta}\right)$

Therefore $h=\frac{\tan \alpha \tan \beta}{\tan \alpha+\tan \beta}($ since $A B=1)$

Hence height of aero plane is $\frac{\tan \alpha \tan \beta}{\tan \alpha+\tan \beta}$