An aeroplane at an altitude of 250 m observes the angle of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. Find the width of the river. Write the answer to the nearest whole number.

Given,

Aeroplane is at point A and boats are at point B and C. Since, aeroplane is at an altitude of 250 m ,

∴ AD = 250 m.

Considering right angled ΔACD, we get

$\Rightarrow \tan \theta = \dfrac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 60^{\circ} = \dfrac{AD}{DC} \\[1em] \Rightarrow \sqrt{3} = \dfrac{250}{y} \\[1em] \Rightarrow y = \dfrac{250}{\sqrt{3}} \\[1em] \Rightarrow y = \dfrac{250}{1.732} \\[1em] \Rightarrow y = 144.34 \text{ m.}$

Considering right angled ΔABD, we get

$\Rightarrow \tan \theta = \dfrac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 45^{\circ} = \dfrac{AD}{BD} \\[1em] \Rightarrow 1 = \dfrac{250}{x} \\[1em] \Rightarrow x = 250 \text{ m.}$

Width of the river (BC) = x + y = 144.34 + 250 = 394.34 meters.

Rounding off to nearest meter BC = 394 meters.

Hence, the width of the river is 394 meters.