Question

The shadow of a tower standing on a level ground is found to be 40 m longer when Suns altitude is $30$ than when it was $60$. Find the height of the tower.

Answer 1

According to Question,
Shadow of tower at ${30}^o$ elevation is $40m$ more than Shadow of tower at ${60}^o$
Let us assume that the Length of Shadow due to ${60}^o$ elevation be $x$

Now, it is Given That $CD=40$
So, in $△ABD$
$\tan {45}^o=\frac{H}{x+40}\Rightarrow H=x+\mathrm{40................}\left(1\right)$
and in $△ABC$
$\tan {60}^o=\frac{H}{x}\Rightarrow \sqrt{3}=\frac{H}{x}\Rightarrow H=x\sqrt{3}.............\left(2\right)$
from (1) and (2)
$x+40=x\sqrt{3}$

$\Rightarrow \sqrt{3}x-x=40$

$\Rightarrow x=\frac{40}{\sqrt{3}-1}=54.8m$ $(Take\sqrt{3}=1.73)$
and $H=40+54.8=94.8m$