Question

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

• A. $20$
• B. $30$
• C. $25$
• D. $40$

Answer 1

The correct option is D $40$

$\theta =\alpha +\beta \Rightarrow \beta =\theta -\alpha$
$\tan \beta =\frac{\tan \theta -\tan \alpha }{1+\tan \theta \tan \alpha }\Rightarrow \frac{3}{5}=\frac{\frac{h}{40}-\frac{h}{160}}{1+\frac{h}{40}.\frac{h}{160}}$
$\Rightarrow h^2-200h+6400=0\Rightarrow h=40$ or $160$