The top of 15 meters high tower makes an angle of depression of 60 degrees with the bottom of a electric pole and an angle of 30 degrees with the top of the pole. What is the height of the electric pole?

AD is the tower = 15 m and CE is the electric pole

Let AB = $$x$$ m and DE = BC = $$y$$ m

Also, $$\angle$$ AED = 60° and $$\angle$$ ACB = 30°

In $$\triangle$$ ADE, => $$tan(\angle AED)=\frac{AD}{DE}$$

=> $$tan(60)=\sqrt{3}=\frac{15}{y}$$

=> $$y=\frac{15}{\sqrt{3}}=5\sqrt{3}$$ m

In $$\triangle$$ ABC, => $$tan(\angle ACB)=\frac{AB}{BC}$$

=> $$tan(30)=\frac{1}{\sqrt{3}}=\frac{x}{5\sqrt{3}}$$

=> $$x=5$$ m

$$\therefore$$ CE = AD - AB = 15 - 5 = 10 meters

=> Ans - (C)