The angle of elevation of the top of a pillar from the foot and the top of a building 20 m high, are 60° and 30° respectively. The height of the pillar is

CE is the building = 20 m and BD = CE = 20 m 

AD is the pillar = ?

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

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

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

=> $$tan(60)=\sqrt{3}=\frac{x+20}{y}$$

=> $$x+20=y\sqrt{3}$$ --------------(i)

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

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

=> $$y=x\sqrt3$$ 

Substituting it in equation (i), we get :

=> $$x+20=(x\sqrt{3}) \times \sqrt3$$

=> $$x+20=3x$$

=> $$3x-x=2x=20$$

=> $$x=\frac{20}{2}=10$$ m

$$\therefore$$ AD = AB + BD = 10 + 20 = 30 m

=> Ans - (D)