The upper $\frac{3}{4}$ th portion of a vertical pole subtends an angle ${\tan }^{-1}\frac{3}{5}$ at point in the horizontal plane through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is:

Let the height of the pole be $H$. The distance from the foot is $d=40$.The pole is divided into lower part $H/4$ and upper part $3H/4$.Let $\alpha$ be the angle subtended by the lower part and $\beta$ by the whole pole.$\tan \alpha =\frac{H/4}{40}=\frac{H}{160}$$\tan \beta =\frac{H}{40}$ The angle subtended by the upper part is $\theta =\beta -\alpha$.Given $\tan \theta =3/5$.
$$\tan (\beta -\alpha )=\frac{\tan \beta -\tan \alpha }{1+\tan \beta \tan \alpha }=\frac{3}{5}$$
$$\frac{H/40-H/160}{1+(H/40)(H/160)}=\frac{3}{5}$$
$$\frac{3H/160}{1+H^2/6400}=\frac{3}{5}$$
$$\frac{H/160}{(6400+H^2)/6400}=\frac{1}{5}$$
$$\frac{40H}{6400+H^2}=\frac{1}{5}⟹200H=6400+H^2$$
$$H^2-200H+6400=0$$
$$(H-160)(H-40)=0$$
$H=40$ or $H=160$.A possible height is 40 m.