Question

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 m from the banks, find the width of the river.

Answer 1

Based upon the given information, we can draw the figure shown above.

In figure, $A$ and $B$ are points on the bank on opposite sides of river.

$\Rightarrow$ $AB$ is the width of the river.

$P$ is the point on bridge at a height $=30m$

$\therefore DP=30m$

Now, $AB=AD+DB$

In right angled $△ADP$,

$\angle A={30}^o$

$\Rightarrow$ $\tan {30}^o=\frac{PD}{AD}$

$\Rightarrow$ $\frac{1}{\sqrt{3}}=\frac{30}{AD}$

$\therefore$ $AD=30\sqrt{3}m$

In right angled $△PDB$,

$\tan {45}^o=\frac{PD}{DB}$

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

$\therefore$ $DB=30m$

Now, $AB=BD+AD=30+30\sqrt{3}m=30+30\left(1.732\right)=81.96m$

$\therefore$ The width of river is $81.96m$.