From the top of a hill  $100m$ high, the angle of depression of the top and bottom of a pole are  $30°$ and  $60°$ respectively.  What is the height of the pole?

Consider $AC$ represents the hill and $DE$ represents the pole

Given that $AC=100m$

$\angle XAD=\angle ADB=30°\left(\because AX\left|\right|BD\right)$

$\angle XAE=\angle AEC=60°\left(\because AX\left|\right|CE\right)$

Let $DE=h$

then $BC=DE=h$

$AB=100-h(\because AC=100,BC=h)$

$BD=CE$

In  $\Delta ACE$

$\tan 60°=\frac{AC}{CE}$

$\Rightarrow \sqrt{3}=\frac{100}{CE}$

$\Rightarrow CE=\frac{100}{\sqrt{3}}\to \left(1\right)$

In $\Delta ABD$, $\tan 30°=\frac{AB}{BD}$

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

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{100-h}{\frac{100}{\sqrt{3}}}(\therefore CE=BD)$

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

$\Rightarrow 100-h=\frac{100}{3}$

$\Rightarrow h=100-33.33$

$\Rightarrow h=66.67m$