From the top of a building 60 metre high, the angle of depression of the top and bottom of a tower are observed to be 30° and 60°. The height of the tower in metre is

Given : AD = 60 m

To find : Height of tower = CE = ?

Solution : By symmetry BD = CE and BC = DE

In $$\triangle$$ ADE,

=> $$tan(\angle AED)=\frac{AD}{DE}$$

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

=> $$\sqrt{3}=\frac{60}{DE}$$

=> $$DE=\frac{60}{\sqrt{3}} = 20\sqrt{3}$$ m

Thus, BC = DE = $$20\sqrt{3}$$ cm

In $$\triangle$$ ABC

=> $$tan(\angle ACB)=\frac{AB}{BC}$$

=> $$tan(30^\circ)=\frac{AB}{20\sqrt{3}}$$

=> $$\frac{1}{\sqrt{3}}=\frac{AB}{20\sqrt{3}}$$

=> $$AB=20$$ m

$$\therefore$$ CE = BD = AD - AB

= $$60-20=40$$ m

=> Ans - (A)