The top of a $15m$ high tower makes an angle of elevation of  $60°$ with the bottom of an electronic pole and angle of elevation of  $30°$ with the top of the pole.  What is the height of the electric pole? (in meters)

$AC$ represents the tower and  $DE$ represents the pole

Given that $AC=15m$

$\angle ADB=30°$

$\angle AEC=60°$

Let $DE=h$, then  $BC=DE=h$

$AB=(15-h)(\because AC=15\text{\hspace{0.17em}and\hspace{0.17em}}BC=h)$

$\tan 60°=\frac{AC}{CE}$

$\Rightarrow \sqrt{3}=\frac{15}{CE}$

$\Rightarrow CE=\frac{15}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$

$\Rightarrow CE=\frac{15\sqrt{3}}{3}$

$\Rightarrow CE=5\sqrt{3}$

$\tan 30°=\frac{AB}{BD}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{15-h}{BD}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{15-h}{5\sqrt{3}}(\because BD=CE)$

$\Rightarrow 5=15-h$

$\Rightarrow h=15-5$

$\Rightarrow h=10m$

$\therefore$ height of the tower is  $10m$.