Question

The shadow of a tower standing on a level ground is found to be $60$ metres longer then the suns altitude is ${30}^o$ than when it is ${45}^o$. Find the height of the tower.

Answer 1

(Refer to Image)

Let height of tower= $AB=x$

$A/Q$ $DC=60m$

In ${\Delta }^{le}ABC,\tan {45}^o=\frac{AB}{BC}$

$\Rightarrow \frac{AB}{BC}=1$

$\Rightarrow BC=x$

In ${\Delta }^{le}ADB$, $\tan {30}^o=\frac{AB}{BD}=\frac{x}{60+x}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{x}{60+x}$

$=\sqrt{x}=60+x$

$\Rightarrow x(\sqrt{3}-1)-60\Rightarrow x=\frac{60}{(\sqrt{3}-1)}$

$\Rightarrow x=81.9615m$