Question

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

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

Answer 1

The correct option is A 40 m

Let $\angle BCD=\alpha$ and $\angle DCA=\beta =tan\frac{3}{5}$
So, $A=\alpha +\beta \Rightarrow \beta =A-\alpha$
$\Rightarrow \tan \beta =\frac{\tan A-\tan \alpha }{1+\tan A\tan \alpha }\Rightarrow \frac{3}{5}=\frac{\frac{h}{40}-\frac{h}{160}}{1+\frac{h}{40}\frac{h}{160}}\Rightarrow h^2-200h+6400=0\Rightarrow (h-40)(h-160)=0$
$\Rightarrow h=40$ or $h=160$