Question

A vertical pole (more than $100\, m$ high) consists of two portions, the lower being one-third 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 40 ft from it, then the height of the pole is:

• A. 100 ft
• B. 120 ft
• C. 150 ft
• D. none of these

Answer 1

Answer: (B)

Let the height of the pole $B C$ be $h$. In $\triangle A B C$,
$\tan \beta=\frac{h}{40}$ ..... (i)

and in $ \Delta \alpha \beta D, $
$ \tan \alpha =\frac{h/3}{40} $
$ =\frac{h}{120} $... (ii)
Now, $ \tan \theta =\frac{1}{2}(given) $
$ \Rightarrow $ $ \tan (\beta -\alpha )=\frac{1}{2} $
$ \Rightarrow $ $ \frac{\tan \beta -\tan \alpha }{1+\tan \beta \tan \alpha }=\frac{1}{2} $
$ \Rightarrow $ $ \frac{\frac{3h}{120}-\frac{h}{120}}{1+\frac{3{{h}^{2}}}{14400}}=\frac{1}{2} $
$ \Rightarrow $ $ h=120,40 $ But $ h $
cannot be taken according to the given condition, therefore $ h=120\,ft. $