From a point on a bridge across a river,the angles of depression of the banks on opposite sides of the river are $30^{\circ}$ and $45^{\circ}$,respectively. If the bridge is at a height of $3 \, m$ from the banks,find the width of the river.

(N/A) Let $A$ and $B$ represent points on the bank on opposite sides of the river,so that $AB$ is the width of the river. Let $P$ be a point on the bridge at a height of $3 \, m$ from the river level,i.e.,$PD = 3 \, m$,where $D$ is the point on the river surface directly below $P$.
In right-angled $\triangle APD$,$\angle PAD = 30^{\circ}$.
Using trigonometry,$\tan 30^{\circ} = \frac{PD}{AD}$.
$\frac{1}{\sqrt{3}} = \frac{3}{AD} \implies AD = 3\sqrt{3} \, m$.
In right-angled $\triangle PBD$,$\angle PBD = 45^{\circ}$.
Using trigonometry,$\tan 45^{\circ} = \frac{PD}{BD}$.
$1 = \frac{3}{BD} \implies BD = 3 \, m$.
The total width of the river is $AB = AD + BD = 3\sqrt{3} + 3 = 3(\sqrt{3} + 1) \, m$.