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)
Create a FREE account and get: