Question

The length of the shadow of a tower standing on level plane is found to be $2x$ meters longer when the sun's altitude is $\text{3}{\text{0}}^{\text{o}}$ than when it was $\text{4}{\text{5}}^{\text{o}}$. Prove that height of tower is $x\left(\sqrt{3}+1\right)$ meters.

Answer 1

In $△BCD$

$\tan {45}^{\circ }=\frac{h}{y}=1$

$\Rightarrow h=y$

In $△ABC$

$\tan {30}^{\circ }=\frac{h}{2x+y}=\frac{1}{\sqrt{3}}$

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

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

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

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

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

But $y=b=x(\sqrt{3}+1)$ meters