From the point on a bridge across a river, the angles of depressions of the banks on opposite sides of the river are 30^o and 45^o, respectively. If the bridge is at a height of 3 m from the banks, find the width of the river.

Let the river width be BD.

Height of bridge = 3 m So, CQ = 3 m

Here, ∠PQB = 30^o (Angle of depression)

∠RQD = 45^o  (Angle of depression)

Since QC is height and it is perpendicular to BD.

Therefore, ∠QCB =∠QCD = 90^o
 

Now, in △QBC,

$\tan {30}^{\circ }\mathrm{ }=\frac{\mathit{QC}}{\mathit{BC}}$

$\frac{1}{\sqrt{3}}=\frac{3}{\mathit{BC}}$

$\mathit{BC}=3\surd 3$ m

Now, in △QCD,

${\tan 45}^{\circ }=\frac{\mathrm{QC}}{\mathrm{CD}}$

$1=\frac{3}{\mathrm{CD}}$

$\mathit{CD}=3$ m

Therefore, $\mathit{BD}=\mathit{BC}+\mathit{CD}$

$=\left(3\sqrt{3}+3\right) \mathrm{m}$

$=3\left(\surd 3+1\right) \mathrm{m}$

Hence, the width of the river is $\mathbf{3}\mathbf{(}\sqrt{\mathbf{3}}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{ }\mathbf{m}.$