Question

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

Answer 1

We are given that $\tan \theta =\frac{1}{2}$
Where $\theta$ is the angle subtended by the upper portion.
Also $\tan \alpha =\frac{3h}{40},\tan \beta =\frac{h}{40}$
and $\theta =\alpha -\beta ,\therefore \tan \theta =\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }$
or $\frac{1}{2}=\frac{\left(3h/40\right)-\left(h/40\right)}{1+\left(3h^2/{40}^2\right)}=\frac{80h}{1600+3h^2}$
or $3h^2+1600=160h$
or $3h^2-160h+1600=0$.
or $3h^2-120h-40h+1600=0$.
or $(h+40)(3h-40)=0$.
$\therefore h=40$ and hence $3h=120$ is the required length of the pole. $3h\ne 40$ as the pole of given to be the more than $100ft.$