Question

Shadow of a tower standing on a level ground is found to be 40 m longer when the sun's altitude is 30 degree than when it is 60 degree. find the height of the tower

Answer 1

In $△ABD\tan {60}^0=\frac{AB}{BD}\sqrt{3}=\frac{h}{x}\therefore x=\frac{h}{\sqrt{3}}.....\left(1\right)$

Now, in $△ABC\tan {30}^0=\frac{AB}{BC}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+40}\Rightarrow x+40=\sqrt{3}h$

$\therefore \frac{h}{\sqrt{3}}+40=\sqrt{3}h$ [From equation (1)]

$\Rightarrow h+40\sqrt{3}=3h\Rightarrow 2h=40\sqrt{3}\therefore h=20\sqrt{3}h$

$\therefore$ The height of a tower is $20\sqrt{3}m$