An aeroplane is flying at a height of 300 m above the ground, Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45 and 30 respectively. Find the width of the river. [Use$\sqrt{3} = 1.732$)

Given: Height of the plane$=300\ m$,  The angles of depression from the airplane of two points on both banks of a river in opposite directions are $45^{o}$ and $30^{o}$ respectively.

To do: To find the width of the river.

Solution:

Given airplane is at height of $300\ m$.

$AB=300\ m$ and $XY || PQ$.

Angles of depression of the two points $P$ & $Q$ are $30^{o}$ and $45^{o}$.

$\angle XAP=30^{o}$ & $\angle YAQ=45^{o}$

$\angle XAP\ =\angle APB=30^{o} \&\ \angle YAQ=\angle AQB=45^{o}$

In $\vartriangle APB$,

$tan 30^{o}=\frac{AB}{PB}$

$\frac{1}{\sqrt{3}} =\frac{AB}{PB}$

$PB=AB\sqrt{3} =300\sqrt{3} \ m$

In $\vartriangle BAQ$,

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

$1=\frac{300}{BQ}$

$BQ=300\ m$

Width of the river $PQ=PB+BQ=300+300\sqrt{3} =300( 1+\sqrt{3}) \ meter$.
 

Updated on: 10-Oct-2022

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