The angles of elevation of the top of a building from the top and bottom of a tree are $$30^\circ$$ and $$30^\circ$$ respectively. If the height of the tree is $$50 m$$, then what is the height of the building?

AD is the building and CE is the tree, thus $$CE=BD= 50$$ m

Let AB = $$x$$ m and DE = BC = $$y$$ m

Also, $$\angle$$ AED = 60° and $$\angle$$ ACB = 30°

In $$\triangle$$ ADE, => $$tan(\angle AED)=\frac{AD}{DE}$$

=> $$tan(60)=\sqrt{3}=\frac{x+50}{y}$$

=> $$y\sqrt{3}=x+50$$

=> $$y=\frac{x+50}{\sqrt3}$$ --------------(i)

In $$\triangle$$ ABC, => $$tan(\angle ACB)=\frac{AB}{BC}$$

=> $$tan(30)=\frac{1}{\sqrt{3}}=\frac{x}{y}$$

=> $$y=x\sqrt3$$

=> $$\frac{x+50}{\sqrt3}=x\sqrt3$$     [Using equation (i)]

=> $$x+50=3x$$

=> $$3x-x=2x=50$$

=> $$x=\frac{50}{2}=25$$

$$\therefore$$ AD = AB + BD = $$x+y=25+50=75$$ m

=> Ans - (B)