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. A possible height of the vertical pole is:

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

Answer 1

The correct option is A 40 m

Here, $\alpha =A-\beta$

$\therefore \beta =A-\alpha$

Hence $\tan \beta =\frac{\tan A+\tan \alpha }{1-\tan A\tan \alpha }$

$\Rightarrow \frac{3}{5}=\frac{\frac{h}{40}+(-\frac{h}{160})}{1-\left(\frac{h}{40}\right)(-\frac{h}{160})}$

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

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

$\Rightarrow h=40orh=160$