Question

The length of the shadow of a tower standing on level plane is found to be $2y$ metres longer when the sun's altitude is ${30}^o$ that when it was ${45}^o$. Prove that the height of the tower is $y(\sqrt{3}+1)$ metres.

Answer 1

$\tan 45=\frac{h}{x}\Rightarrow h=x$...(1)

$\tan 30=\frac{h}{2y+x}$

$\frac{1}{\sqrt{3}}=\frac{h}{2y+x}$

$2y+x=\sqrt{3}h$

$2y+h=\sqrt{3}h$

$2y=(\sqrt{3}-1)h$

$h=\frac{2y}{\sqrt{3}-1}$

$h=\frac{2y}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

$h=(\sqrt{3}+1)y$