The shadow of a tree when the angle of elevation of the sun at $$45^{\circ}$$ is found to be $$20\ m$$ longer than when it is $$60^{\circ}$$. Find the height of the tree. $$(\sqrt {3} = 1.73)$$.

In $$\triangle ABC, \dfrac {h}{x} = \tan 60^{\circ}$$
$$\Rightarrow h = x\tan 60^{\circ} .... (!)$$
In $$\triangle ABD, \dfrac {h}{x + 20} = \tan 45^{\circ}$$
$$\therefore \dfrac {x\tan 60^{\circ}}{x + 20} = \tan 45^{\circ}$$ [From equation (1)]
$$\Rightarrow \sqrt {3}x = 1\times (x + 20)\Rightarrow \sqrt {3} x = x + 20$$
$$\Rightarrow 1.73x - x = 20$$
$$\Rightarrow x = \dfrac {20}{0.73} = 27.39$$
Hence, height of tree $$= h = 27.39 \times \sqrt {3} = 47.39\ m$$.