An aeroplane is flying at a height of $210\ m$. Flying at this height at some instant the angles of depression of two points in a line in opposites directions on both the banks of the river are $45^{\circ}$ and $60^{\circ}$. Find the width of the river. (Use $\sqrt3=1.73$)

Given:

An aeroplane is flying at a height of $210\ m$. Flying at this height at some instant the angles of depression of two points in a line in opposites directions on both the banks of the river are $45^{\circ}$ and $60^{\circ}$. 

To do:

We have to find the width of the river.

Solution:

Let the aeroplane is at a height of $AB=210\ m$. Let $C$ and $D$ be the two points in a line in opposites directions on both the banks of the river.

Angles of depression of the two points $C$ and $D$ are $60^{o}$ and $45^{o}$.

$\angle BCA=60^{o}$ and $\angle BDA=45^{o}$

In $\vartriangle BCA$,

$tan 60^{o}=\frac{AB}{CA}$

$\sqrt{3}=\frac{210}{CA}$

$CA=\frac{210}{\sqrt3} \ m$

In $\vartriangle BAD$,

$tan45^{o}=\frac{AB}{AD}$

$1=\frac{210}{AD}$

$AD=210\ m$

Therefore,

The width of the river $CD=CA+AD=210+\frac{210}{\sqrt3}$

$=\frac{210\sqrt3+210}{\sqrt3}\ m$

$=\frac{210(1.73+1)}{1.73}\ m$

$=\frac{210(2.73)}{1.73}\ m$

$=\frac{573.3}{1.73}\ m$

$=331.38\ m$

Therefore, the width of the river is $331.38\ m$.

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Updated on: 10-Oct-2022

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