A man standing on the bank of river observes that the angle subtended by a tree on the opposite bank is 60o. When he retires 36 m from the bank, he finds that the angle is 30o. The breadth of the river is

Given : CD is the tree = $$h$$ m and AB = 36 m

To find : DB = $$x$$ = ?

Solution : In $$\triangle$$ BCD,

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

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

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

Again, in $$\triangle$$ ACD,

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

=> $$\frac{1}{\sqrt{3}}=\frac{x\sqrt3}{x+36}$$     [Using (i)]

=> $$x+36=(x\sqrt{3})\times \sqrt3=3x$$

=> $$3x-x=2x=36$$

=> $$x=\frac{36}{2}=18$$ m

=> Ans - (B)