The elevation of the top of a tower from a point on the ground is $$\ 45^\circ$$. On travelling 60 m from the point towards the tower, the alevation of the top becomes $$\ 60^\circ$$. The height of the tower, in metres, is

From $$\triangle$$ ACD,

tan $$60^\circ = \frac{AD}{CD}$$

$$\Rightarrow AD = CD\sqrt{3}$$

From $$\triangle$$ ABD,

tan $$45^\circ = \frac{AD}{BD}$$

$$\Rightarrow$$ AD = BD

$$\Rightarrow$$ AD = BC+CD

$$\Rightarrow$$ $$AD = 60+\frac{AD}{\sqrt{3}}$$

$$\Rightarrow$$ $$AD-\frac{AD}{\sqrt{3}} = 60$$

$$\Rightarrow \frac{\sqrt{3}AD-AD}{\sqrt{3}} = 60$$

$$AD(\sqrt{3}-1) = 60\sqrt{3}$$

$$AD = \frac{60\sqrt{3}}{\sqrt{3}-1}$$

Rationalising above equation

$$AD = \frac{60\sqrt{3}}{\sqrt{3}-1}\times\frac{\sqrt{3}+1}{\sqrt{3}+1}$$

$$AD = \frac{60\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}}$$

$$\therefore AD = 30(\sqrt{3}+3)$$