The upper $\frac{3}{4}$ th portion of a vertical pole subtends an angle $\text{'}\theta \text{'}$ such that $\tan \theta =\frac{3}{5}$ at a point in the horizontal plane through its foot and at a distance 40m from the foot.  Find the possible height of the vertical pole

Given $\tan \theta =\frac{3}{5}$

$PA=40m$

Let $\therefore AC=h$ then

$\begin{array}{l}BC=\frac{3}{4}h\\ AB=\frac{1}{4}h\\ \angle APB=\alpha \end{array}$

In $\Delta APB$,

$\begin{array}{l}\tan \alpha =\frac{AB}{AP}\Rightarrow \tan \alpha =\frac{\frac{1}{4}h}{40}\Rightarrow \frac{h}{160}=\tan \alpha \\ h=160\tan \alpha \end{array}$

In $\Delta APC$

$\begin{array}{l}\tan (\theta +\alpha )=\frac{h}{40}\Rightarrow h=40\tan (\theta +\alpha )\\ \Rightarrow \tan (\theta +\alpha )=\frac{h}{40}\\ \Rightarrow \frac{\tan \theta +\tan \alpha }{1-\tan \theta \cdot \tan \alpha }=\frac{h}{40}\end{array}$

$\begin{array}{l}\Rightarrow \frac{\frac{3}{5}+\frac{h}{160}}{1-\frac{3}{5}\times \frac{h}{160}}=\frac{h}{40}\Rightarrow \frac{480+5h}{800-3h}=\frac{h}{40}\\ \Rightarrow 19200+200h=800h-3h^2\end{array}$

$\begin{array}{l}\Rightarrow 3h^2-600h+19200=0\\ \Rightarrow h^2-200h+6400=0\\ \Rightarrow h^2-160h-40h+6400=0\\ \Rightarrow (h-160)(h-40)=0\end{array}$

$\therefore h=160$ (or) $h=40$