A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. Find :

(i) the height of the tree, correct to 2 decimal places.

(ii) the width of the river.

Let CD be the tree and B be the position of the person when angle of elevation is 60° and A be the position when angle of elevation is 30°.

(i) In △BCD,

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

In △ACD,

$\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{CD}{AC} \\[1em] \Rightarrow AC = CD\sqrt{3}.$

From figure,

AB = AC - BC

$\Rightarrow 40 = CD\sqrt{3} - \dfrac{CD}{\sqrt{3}} \\[1em] \Rightarrow \dfrac{3CD - CD}{\sqrt{3}} = 40 \\[1em] \Rightarrow \dfrac{2CD}{\sqrt{3}} = 40 \\[1em] \Rightarrow CD = \dfrac{40\sqrt{3}}{2} = 20\sqrt{3} \\[1em] \Rightarrow CD = 20 \times 1.732 = 34.64 \text{ m}.$

Hence, height of tree = 34.64 meters.

(ii) From part (i), we get :

BC = $\dfrac{CD}{\sqrt{3}} = \dfrac{20\sqrt{3}}{\sqrt{3}}$

= 20 meters.

Hence, width of river = 20 meters.