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 correct to the nearest whole number.

Let aeroplane be at point D and boats be at point A and B. Since, aeroplane is at an altitude of 250 m therefore,

∴ CD = 250 m.

From figure,

∠DAC = ∠XDA = 45° (Alternate angles are equal)
∠DBC = ∠YDB = 60° (Alternate angles are equal)

Considering right angled △BCD, we get

$\Rightarrow \text{tan 60°} = \dfrac{CD}{BC} \\[1em] \Rightarrow \sqrt{3} = \dfrac{250}{BC} \\[1em] \Rightarrow BC = \dfrac{250}{\sqrt{3}} = 144.34$

Considering right angled △ACD, we get

$\Rightarrow \text{tan 45°} = \dfrac{CD}{AC} \\[1em] \Rightarrow 1 = \dfrac{250}{AC} \\[1em] \Rightarrow AC = 250$

Width of the river (AB) = AC + BC = 144.34 + 250 = 394.34 meters.

Rounding off to nearest meter AB = 394 meters.

Hence, the width of the river is 394 meters.