Question

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

• A. $40m$ or $100m$
• B. $160m$ or $40m$
• C. $100m$ or $120m$
• D. $200m$ or $80m$

Answer 1

Answer: (B)

$160m$ or $40m$

Given, ${\theta }_2={\tan }^{-1}\frac{3}{5}\Rightarrow \tan {\theta }_2=\frac{3}{5}$ ...(i)

In $△AOC,\tan {\theta }_1=\frac{AC}{AO}=\frac{h}{160}$ ...(ii)

and in $△AOB,\tan ({\theta }_1+{\theta }_2)=\frac{h}{40}$

$\Rightarrow \frac{\tan {\theta }_1+\tan {\theta }_2}{1-\tan {\theta }_1\tan {\theta }_2}=\frac{h}{40}$

$\Rightarrow \frac{\frac{h}{160}+\frac{3}{5}}{1-\frac{h}{160}\times \frac{3}{5}}=\frac{h}{40}$ [from Eqs. (i) and (ii)]

$\Rightarrow \frac{5(h+96)}{800-3h}=\frac{h}{40}$

$\Rightarrow h^2-200h+6400=0$

$\Rightarrow (h-160)(h-40)=0$

$\Rightarrow h=160orh=40$

Hence, the height of the vertical pole is $40m$