A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60°. When he moves 50 m away from the bank, he finds the angle of elevation to be 30°. Calculate :

(i) the width of the river and

(ii) the height of the tree.

(i) Let AB be the width of the river and BC be the tree.

In △ABC,

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

In △DBC,

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

From (1) and (2) we get,

$\Rightarrow \sqrt{3}AB = \dfrac{BD}{\sqrt{3}} \\[1em] \Rightarrow 3AB = BD \\[1em] \Rightarrow 3AB = AD + AB \\[1em] \Rightarrow 3AB - AB = AD \\[1em] \Rightarrow 2AB = AD \\[1em] \Rightarrow 2AB = 50 \\[1em] \Rightarrow AB = \dfrac{50}{2} \\[1em] \Rightarrow AB = 25 \text{ m}.$

Hence, the width of river is 25 metres.

(ii) From equation (1),

BC = $\sqrt{3}$AB = 1.732 × 25 = 43.3 meters.

Hence, height of tree = 43.3 meters.