Question

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

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

Answer 1

Answer: (B)

$tan\left(\theta + \alpha \right)=\frac{x}{40}$
$\&tan\theta =\frac{x}{160}$
$tan\alpha =\frac{3}{5}$
$\therefore \alpha =\left(\right.\left(\right.\theta +\alpha \left.\right)-\theta \left.\right)$
$\Rightarrow tan\alpha =tan\left(\right.\left(\right.\theta +\alpha \left.\right)-\theta \left.\right)$
$\Rightarrow \frac{3}{5}=\frac{\frac{x}{40} - \frac{x}{160}}{1 + \frac{x}{40} \cdot \frac{x}{160}}$
$\therefore x=40\,m$