The angle of elevation of the top of a tower from a point A on the ground is 30˚. On moving a distance of 20 metres towards the foot of the tower to a point B, the angle of elevation increases to 60˚. The height of the tower in metres is

Given : CD is the tower and AB = 20 m

To find : CD = $$h$$ = ?

Solution : Let DB = $$x$$ m

In $$\triangle$$ BCD,

=> $$tan(60^\circ)=\frac{CD}{DB}$$

=> $$\sqrt{3}=\frac{h}{x}$$

=> $$x=\frac{h}{\sqrt{3}}$$ -----------(i)

Again, in $$\triangle$$ ACD,

=> $$tan(30^\circ)=\frac{CD}{AD}$$

=> $$\frac{1}{\sqrt{3}}=\frac{h}{x+20}$$

=> $$x+20=\sqrt{3}h$$

Substituting value of $$x$$ from equation (i), we get :

=> $$\sqrt{3}h-\frac{h}{\sqrt{3}}=20$$

=> $$\frac{2h}{\sqrt{3}}=20$$

=> $$h=20 \times \frac{\sqrt{3}}{2}$$

=> $$h=10\sqrt{3}$$ m

=> Ans - (C)