Asked in: July 2024 board paperTo find the width of the river, a man observes the top of a tree on the opposite bank making an angle of elevation of 60°. When he moves 24 metre backward from the bank and observes the same top of the tree, his line of…

Answer: Width of river = 12 m; height of tree \(= 12\sqrt3 = 20.76\) m

[Figure: Tree AB on the far bank, B its foot. The man first stands at C on the near bank, then at D with DC = 24 m; D, C, B in a line. \(\angle ACB = 60^\circ\), \(\angle ADB = 30^\circ\).]

Let height AB = h and width of river BC = x.

In right-angled ΔABC: \(\tan 60^\circ = \dfrac{h}{x}\) \(\therefore h = \sqrt3\,x\) ...(I)

In right-angled ΔABD: \(\tan 30^\circ = \dfrac{h}{x + 24}\) \(\therefore h = \dfrac{x + 24}{\sqrt3}\) ...(II)

From (I) and (II): \(\sqrt3\,x = \dfrac{x + 24}{\sqrt3}\) \(\therefore 3x = x + 24\) \(\therefore 2x = 24\) \(\therefore x = 12\)

\(h = 12\sqrt3 = 12 \times 1.73 = 20.76\)

\(\therefore\) Width of the river is 12 m and height of the tree is 20.76 m.