Question

The shadow of a tower, when the angle of elevation of the sun is ${45}^o$, is found to be $10$ m. longer than when it was ${60}^o$. Find the height of the tower.

Answer 1

According to Question,
Shadow of tower at ${45}^o$ elevation is $10m$ 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 $CD=10$
So, in $△ABD$
$\tan {45}^o=\frac{H}{x+10}\Rightarrow H=x+\mathrm{10................}\left(1\right)$
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)
$x+10=x\sqrt{3}\Rightarrow \sqrt{3}x-x=10$

$\Rightarrow x=\frac{10}{\sqrt{3}-1}=13.69m$ $(Take\sqrt{3}=1.73)$
and $H=10+13.69=23.69m$