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 $=6 \mathrm{~m}$