Question

The upper $3/4^{th}$ portion of a vertical pole subtends an angle $ta{n}^{-1}\frac{3}{5}$ at a point in the horizontal plane through its foot and at a distance $40$ m from its foot. Given the vertical pole is at height less than $100$ m from the ground. Find its height.

Answer 1

Given that,

$\alpha ={\tan }^{-1}\frac{3}{5}\tan \alpha =\frac{3}{5}$

And we know that In the above figure,

$\tan \beta =\frac{AC}{OA}=\frac{h}{4\times 40}=\frac{h}{160}$

Now, In $\Delta AOB,$

$\tan (\alpha +\beta )=\frac{AB}{OA}=\frac{h}{40}$

$\Rightarrow \tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$

$\Rightarrow \frac{h}{40}=\frac{\frac{h}{160}+\frac{3}{5}}{1-\frac{h}{160}\times \frac{3}{5}}$

$\Rightarrow \frac{h}{40}=\frac{\frac{h+96}{160}}{\frac{800-3h}{800}}$

$\Rightarrow \frac{h}{40}=\frac{h+96}{160}\times \frac{800}{800-3h}$

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

$\Rightarrow 800h-3h^2=200(h+96)$

$\Rightarrow 800h-3h^2=200h+19200$

$\Rightarrow 3h^2-600h+19200=0$

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

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

$\Rightarrow h-160=0,h-40=0$

$\Rightarrow h=160andh=40$

Hence $h=40$m. be the height of pole and less then $100$m.