Question

The angle of elevation of the top of a rock from the top and foot of a $100m$ high tower are respectively ${30}^o$ and ${45}^o$. Find the height of the rock.

Answer 1

In $△ABD,\angle ADB={45}^o$

$\therefore \tan 45=\frac{AB}{BD}$

$1=\frac{h}{BD}$

$BD=h$

In $△AEC,\angle ACE={30}^o$

$\therefore \tan 30=\frac{AE}{CE}$

$\frac{1}{\sqrt{3}}=\frac{h-100}{h}$

$h=\sqrt{3}h-100\sqrt{3}$

$\sqrt{3}h-h=100\sqrt{3}$

$h(\sqrt{3}-1)=100\sqrt{3}$

$h=\frac{100\sqrt{3}}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

$=\frac{100\sqrt{3}}{3-1}$

$=50\sqrt{3}$