PQ is a post of given height a, and AB is a tower at some distance.

Solution:

Let AB be the tower of height H and PQ is a given post of height a , and  are angles of elevation of top of tower from P and Q. Let PA = x. and BC = h.

The corresponding figure is as follows

In ∆QCB, 

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

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

Again in ∆PAB,

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

$\Rightarrow \quad \tan \alpha=\frac{(h+a) \tan \beta}{h}$

$\Rightarrow \quad h \tan \alpha=(h+a) \tan \beta$

$\Rightarrow h(\tan \alpha-\tan \beta)=a \tan \beta$

$\Rightarrow \quad h=\frac{a \tan \beta}{\tan \alpha-\tan \beta}$

Now

$\Rightarrow \quad x=\frac{a \tan \beta}{(\tan \alpha-\tan \beta) \times \tan \beta}$

$\Rightarrow \quad x=\frac{a}{\tan \alpha-\tan \beta}$

$\Rightarrow \quad H=a+\frac{a \tan \beta}{\tan \alpha-\tan \beta}$

$\Rightarrow \quad H=\frac{a \tan \alpha}{\tan \alpha-\tan \beta}$

Hence required height is $\frac{a \tan \alpha}{\tan \alpha-\tan \beta}$. And distance is $\frac{a}{\tan \alpha-\tan \beta}$