Question

The length of the shadow of a tower standing on level planes is found to be $2x$ metres longer when the suns altitude is ${30}^{\circ }$ than when it was ${45}^{\circ }$. Prove that the height of tower is $x(\sqrt{3}+1)$ metres.

Answer 1

Here, $h$ is the height of the tower.

$\Rightarrow$ In $△BCD,\tan {45}^o=\frac{DC}{BC}$

$\Rightarrow$ $1=\frac{h}{y}$

$\therefore$ $h=y$ ------- ( 1 )

$\Rightarrow$ In $△ABC,\tan {30}^o=\frac{h}{2x+y}$

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

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

$\Rightarrow$ $2x+h=\sqrt{3}h$ [ From( 1 ) ]

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

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

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

$\therefore$ $h=x(\sqrt{3}+1)$ [ Hence proved ]