Given -

Distance AB= 50M

Bearing of A =48$^\circ$

Bearing of B=348$^\circ$

Calculate the width of the river CD?

$\angle A=90^\circ-48^\circ=42^\circ \\ 360^\circ-348^\circ=12^\circ \\ \angle A=90^\circ-12^\circ=78^\circ $

$In\ \Delta ADC, \tan42^\circ=\frac{CD}{AD} $ $$AD=\frac{CD}{\tan42^\circ}\tag{1}$$

$In\ \Delta BDC, \tan78^\circ=\frac{CD}{BD} $ $$BD=\frac{CD}{\tan78^\circ}\tag{2}$$

From equation 1 and 2

$$AD+BD=\frac{CD}{\tan42^\circ}+\frac{CD}{\tan78^\circ} \\ 50=\frac{CD}{0.90}+\frac{CD}{4.70} \\ CD=37.76\ m$$ …