From a point on a bridge across a river the angles of depression of the banks on opposite side of the river are $ 30^{\circ} $ and $ 45^{\circ} $ respectively. If bridge is at the height of $ 30 \mathrm{~m} $ from the banks, find the width of the river.

Given:

From a point on a bridge across a river the angles of depression of the banks on opposite side of the river are \( 30^{\circ} \) and \( 45^{\circ} \) respectively.

The bridge is at the height of \( 30 \mathrm{~m} \) from the banks.

To do:

We have to find the width of the river.

Solution:  

Let $AB$ be the height of the bridge and $C, D$ are the points of depression of the banks on the opposite side of the river.

From the figure,

$\mathrm{AB}=30 \mathrm{~m}, \angle \mathrm{FAD}=\angle \mathrm{ADB}=30^{\circ}, \angle \mathrm{EAC}=\angle \mathrm{BCA}=45^{\circ}$

Let the distance between the bridge and point $C$ be $\mathrm{BC}=x \mathrm{~m}$ and the distance between the bridge and point $D$ be $\mathrm{BD}=y \mathrm{~m}$.

We know that,

$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$

$=\frac{\text { AB }}{BC}$

$\Rightarrow \tan 45^{\circ}=\frac{30}{x}$

$\Rightarrow 1=\frac{30}{x}$

$\Rightarrow x=30 \mathrm{~m}$.........(i)

Similarly,

$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$

$=\frac{\text { AB }}{BD}$

$\Rightarrow \tan 30^{\circ}=\frac{30}{y}$

$\Rightarrow \frac{1}{\sqrt3}=\frac{30}{y}$           

$\Rightarrow y=30\sqrt3 \mathrm{~m}$..........(ii)

$\Rightarrow x+y=30+30\sqrt3=30(\sqrt3+1) \mathrm{~m}$            

Therefore, the width of the river is $30(\sqrt3+1) \mathrm{~m}$.  

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Updated on: 10-Oct-2022

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