Find the height of a tree when it is found that on walking away from it 20 m, in a horizontal line through its base, the elevation of its top changes from 60° to 30°.

Let AB be the tree.

Let the two points be C and D such that CD = 20 m, ∠ADB = 30° and ∠ACB = 60°.

In ∆ABC,

$\Rightarrow \text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{AB}{BC} \\[1em] \Rightarrow AB = BC\sqrt{3} ……….(1)$

In ∆ABD,

$\Rightarrow \text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{BD} \\[1em] \Rightarrow AB = \dfrac{BD}{\sqrt{3}} ……….(2)$

From (1) and (2), we get :

$\Rightarrow BC\sqrt{3} = \dfrac{BD}{\sqrt{3}} \\[1em] \Rightarrow BD = 3BC \\[1em] \Rightarrow BC + CD = 3BC \\[1em] \Rightarrow CD = 2BC \\[1em] \Rightarrow BC = \dfrac{CD}{2} \\[1em] \Rightarrow BC = \dfrac{20}{2} \\[1em] \Rightarrow BC = 10 \text{ meters}.$

AB = $BC \times \sqrt{3}$ = BC × 1.732 = 17.32 meters.

Hence, the height of the tree is 17.32 metres.