A vertical pole (more than 50 meters high) consists of two portions. The lower being  $\frac{1}{3}rd$of the whole. If the upper portion subtends an angle  ${\tan }^{-1}\left(\frac{1}{2}\right)$at a point in a horizontal plane through the foot of the pole at a distance 40 meters from it, then the height of the pole is

Let AB = h meter, BC = 2h meter

$\angle APC=\alpha ,\angle APB=\beta ,\angle BPC=\theta$ so that $\theta =\alpha -\beta$

It is given that $\tan \theta =\frac{1}{2}$

and $AP=40m$ , where we have $\tan \alpha =\frac{3h}{40},\tan \beta =\frac{h}{40}$

$Now \tan \theta =\tan \left(\alpha -\beta \right)=\frac{\tan \alpha -\tan \beta }{1+\tan \alpha +\tan \beta }$

$\Rightarrow \frac{1}{2}=\frac{\frac{3h}{40}-\frac{h}{40}}{1+\left(\frac{3h}{40}\right)\left(\frac{h}{40}\right)}=\frac{80h}{1600+3h^2}$

$\Rightarrow 3h^2-160h+1600=0\Rightarrow \left(3h-40\right)\left(h-40\right)=0$

$\Rightarrow h=40$ , and hence $3h=120$ is the required length of the pole. Note that  as the pole is given to $3h\ne 40$ be more than 50 meters height.