Let $A B$ be the tower and $C$ and $D$ be two points such that $A C=4 \mathrm{~m}$ and $A D=9 m$. Let:

$\mathrm{AB}=\mathrm{hm}, \angle \mathrm{BCA}=\theta$ and $\angle \mathrm{BDA}=90^{\circ}-\theta$

In the right $\triangle \mathrm{BCA}$, we have:

$\tan \theta=\frac{\mathrm{AB}}{\mathrm{AC}}$

$\Rightarrow \tan \theta=\frac{\mathrm{h}}{4}$

In the right $\triangle \mathrm{BDA}$, we have:

$\tan \left(90^{\circ}-\theta\right)=\frac{\mathrm{AB}}{\mathrm{AD}}$

$\Rightarrow \cot \theta=\frac{\mathrm{h}}{\mathrm{g}} \quad\left[\tan \left(90^{\circ}-\theta\right)=\cot \theta\right]$

$\Rightarrow \frac{1}{\tan \theta}=\frac{h}{9}$….2

$\left[\cot \theta=\frac{1}{\tan \theta}\right]$

Multiplying equations (1) and (2), we get

$\tan \theta \times \frac{1}{\tan \theta}=\frac{\mathrm{h}}{4} \times \frac{\mathrm{h}}{9}$

$\Rightarrow 1=\frac{\mathrm{h}^{2}}{36}$

$\Rightarrow 36=\mathrm{h}^{2}$

$\Rightarrow h=\pm 6$

Height of a tower cannot be negative

$\therefore$ Height of the tower $=6 \mathrm{~m}$