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 $40m$ from the foot. A possible height of the vertical pole is :

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

Answer 1

Answer: (B)

Given ${\theta }_2={\tan }^{-1}\frac{3}{5}$

$\tan {\theta }_2=\frac{3}{5}$
ln $\Delta AOC,\tan {\theta }_1=\frac{AC}{OA}$
$=\frac{\frac{1}{4}h}{40}=\frac{h}{160}$
In $\Delta AOB$,
$\tan \left({\theta }_1+{\theta }_2\right)=\frac{AB}{OA}=\frac{h}{40}$
$\frac{\tan {\theta }_1+\tan {\theta }_2}{1-\tan {\theta }_1\tan {\theta }_2}=\frac{h}{40}$
$\Rightarrow \frac{\frac{h}{160}+\frac{3}{5}}{1-\frac{h}{160}\times \frac{3}{5}}=\frac{h}{40}$
$\frac{5[h+96]}{800-3h}=\frac{h}{40}$
$\Rightarrow 200[h+96]=800h-3h^2$
$\Rightarrow 200h+19200=800h-3h^2$
$\Rightarrow 3h^2-600h+19200=0$
$\Rightarrow h^2-200h+6400=0$
$\Rightarrow (h-160)(h-40)=0$
$\Rightarrow h=160$ or $h=40$
$\therefore$ Height of the vertical pole $=40m$.