From the top of a light-house 60 m high with its base at the sea-level, the angle of depression of a boat is $$15^\circ$$. The distance of the boat from the foot of the light-house is

Given : AB = 60 m and $$\angle$$ ACB = $$15^\circ$$

To find : BC = ?

Solution : Using, $$tan(A-B)=\frac{tan(A)-tan(B)}{1+tan(A)tan(B)}$$

=> $$tan(15^\circ)=tan(60^\circ-45^\circ)$$

= $$\frac{tan(60^\circ)-tan(45^\circ)}{1+tan(60^\circ)tan(45^\circ)}$$

= $$\frac{\sqrt3-1}{1+\sqrt3}$$ ---------------(i)

In right $$\triangle$$ ABC,

=> $$tan(15^\circ)=\frac{AB}{BC}$$

=> $$\frac{\sqrt3-1}{\sqrt3+1}=\frac{60}{BC}$$

=> $$BC=60(\frac{\sqrt3+1}{\sqrt3-1})$$ m