Question

An aeroplane at an altitude of 250m observes the angle of depression of two boats on opposite banks of a river to be ${45}^o$ and ${60}^o$ respectively. Find width of the river.
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• A. $250(1+\frac{1}{\sqrt{3}})m$
• B. $250(1-\frac{1}{\sqrt{3}})m$
• C. $250(\sqrt{3}+\frac{1}{\sqrt{3}})m$
• D. $250(\sqrt{3}-\frac{1}{\sqrt{3}})m$

Answer 1

The correct option is A $250(1+\frac{1}{\sqrt{3}})m$

Given:
Aeroplane is a height of 250 m.
Angle of depression on bank 1 = ${45}^o$
Angle of depression on bank 2 = ${60}^o$

To find: Width of the river (BC).
Trigonometric ratio connecting opposite side and adjacent side is $tan\theta$.

In triangle ABD,
$tan{45}^o=\frac{AD}{BD}=\frac{250}{BD}$
$1=\frac{250}{BD}(\because tan{45}^o=1)$
$BD=250m$

In triangle ADC,
$tan{60}^o=\frac{AD}{DC}=\frac{250}{DC}$
$\sqrt{3}=\frac{250}{DC}(\because tan{60}^o=\sqrt{3})$
$DC=\frac{250}{\sqrt{3}}m$

From the figure we know that,
$BC=BD+DC=250+\frac{250}{\sqrt{3}}=250(1+\frac{1}{\sqrt{3}})m$

Width of the river is $250(1+\frac{1}{\sqrt{3}})m$.