Question

The shadow of the vertical tower on level ground increases by $10$ meters,when the altitude of the sun changes from angle of elevation $45$ to $30$ degrees.Find the height of the tower,correct to one place of decimal. (take $\sqrt{3}=1.73$)

Answer 1

Let the height be $h$

In $△ABC,\tan {45}^{\circ }=\frac{AB}{BC}$

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

$\Rightarrow x=h$

In $△ABD,\tan {30}^{\circ }=\frac{AB}{BD}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+10}$

$\Rightarrow h+10=h\sqrt{3}$

$\Rightarrow h(\sqrt{3}-1)=10$

$\therefore h=\frac{10}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{10(\sqrt{3}+1)}{3-1}=\frac{10(\sqrt{3}+1)}{2}=5(\sqrt{3}+1)$m