Question

The upper $3 / 4$ th portion of a vertical pole subtends an angle $\tan ^{-1} 3 / 5$ at a 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 :

• A. $20 m$
• B. $40 m$
• C. $60 m$
• D. $80 m$

Answer 1

Answer: (B)

Given $\theta_{2}=\tan ^{-1} \frac{3}{5}$

$ \tan \theta_{2} =\frac{3}{5} $
ln $\Delta A O C, \tan \theta_{1} =\frac{A C}{O A} $
$=\frac{\frac{1}{4} h}{40}=\frac{h}{160} $
In $\Delta A O B$,
$\tan \left(\theta_{1}+\theta_{2}\right)=\frac{A B}{O A}=\frac{h}{40}$
$\frac{\tan \theta_{1}+\tan \theta_{2}}{1-\tan \theta_{1} \tan \theta_{2}}=\frac{h}{40}$
$\Rightarrow \frac{\frac{h}{160}+\frac{3}{5}}{1-\frac{h}{160} \times \frac{3}{5}}=\frac{h}{40}$
$\frac{5[h+96]}{800-3 h}=\frac{h}{40}$
$\Rightarrow 200[h+96]=800 h-3 h^{2}$
$\Rightarrow 200 h+19200=800 h-3 h^{2} $
$ \Rightarrow 3 h^{2}-600 h+19200=0 $
$\Rightarrow h^{2}-200 h+6400=0 $
$ \Rightarrow (h-160)(h-40)=0 $
$ \Rightarrow h=160 $ or $ h=40$
$\therefore $ Height of the vertical pole $=40 m$.