Question

From a point on a bridge across a river the angles of depression of the banks on opposite sides of the river are 30${}^o$ and 45${}^o$ respectively. If the bridge is at a height of 4 m from the banks, find the width of the river.

Answer 1

The given information is drawn in the attached figure.

Given:

$AB(=CD=EF)=4m$

$\angle FAE={30}^o$

$\angle DAC={45}^o$

From definition of tan of an angle $\theta$,we know: $\tan \left(\theta \right)=\frac{\text{Perpendicular}}{\text{Base}}$

Hence, $\tan \left(\angle FAE\right)=\frac{FE}{FA}$

$\tan \left({30}^o\right)=\frac{4}{FA}$

$\frac{1}{\sqrt{3}}=\frac{4}{FA}$

$FA=4\sqrt{3}m$

Similarly, $\tan \left(\angle DAC\right)=\frac{CD}{DA}$

$\tan \left({45}^o\right)=\frac{4}{DA}$

$1=\frac{4}{DA}$

$DA=4m$

Width of the river is: $w=EC$

$w=FD$

$w=FA+AD$

$w=4\sqrt{3}+4m$