The angles of elevation of the top of a temple, from the foot and the top of a building 30 m high, are 60° and 30° respectively. Then height of the temple is

CE is the building = 30 m and AD is the temple

By symmetry, BD = CE = 30 m and  DE = BC = $$y$$ m

Let AB = $$x$$ m

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

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

=> $$tan(60)=\sqrt{3}=\frac{x+30}{y}$$

=> $$x+30=y\sqrt{3}$$ ------------(i)

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

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

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

Substituting it in equation (i), we get :

=> $$x+30=(x\sqrt{3}) \times \sqrt{3}=3x$$

=> $$3x-x=2x=30$$

=> $$x=\frac{30}{2}=15$$ m

$$\therefore$$ AD = AB + BD = 15 + 30 = 45 meters

=> Ans - (D)