Let a vertical tower AB have its end . A  on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If $\mathrm{\angle }BPC=\beta ,\text{ then }\tan \beta$ is equal to.

Let $AB=h,\text{ then }AD=2h\text{ and }AC=BC=\frac{h}{2}$

Again let, $\mathrm{\angle }CPA=\alpha$

Now, $\mathrm{\Delta }ABP,\tan (\alpha +\beta )=\frac{AB}{AP}=\frac{h}{2h}=\frac{1}{2}$

Also in, $\mathrm{\Delta }ACP,\tan \alpha =\frac{AC}{AP}=\frac{\frac{h}{2}}{2h}=\frac{1}{4}$

Now,        $\tan \beta =\tan \left[\right(\alpha +\beta )-\alpha ]$

$\begin{array}{r}=\frac{\tan (\alpha +\beta )-\tan \alpha }{1+\tan (\alpha +\beta )\tan \alpha }\\ =\frac{\frac{1}{2}-\frac{1}{4}}{1+\frac{1}{2}\times \frac{1}{4}}=\frac{\frac{1}{4}}{\frac{9}{8}}=\frac{2}{9}\end{array}$