Given, Speed of the river (${\mathrm{v}}_{\mathrm{r}}$) = 3m/s (east)
Speed of swimmer (${\mathrm{v}}_{\mathrm{s}}$) = 4 m/s (east)

a) When the swimmer starts swimming due north then his resultant velocity

$\begin{array}{l}\mathrm{v}=\sqrt{{\mathrm{v}}_{\mathrm{r}}^2+{\mathrm{v}}_{\mathrm{s}}^2}=\sqrt{(3{)}^2+(4{)}^2}\\ =\sqrt{9+16}=\sqrt{25}=5\mathrm{m}/\mathrm{s}\\ \mathrm{tan\theta } = \frac{{\mathrm{v}}_{\mathrm{t}}}{{\mathrm{v}}_{\mathrm{s}}} = \frac{3}{4}\\ = 0.75 = \tan {36}^{\circ }54\text{'}\\ \mathrm{Hence}, \mathrm{\theta }={36}^{\circ }54\text{'}\mathrm{N}\end{array}$

Step 2: Find the velocity of the swimmer so that he reaches directly opposite point.

(b) To reach opposite points B, the swimmer should swim at an angle $\mathrm{\theta }$ of the north. Resultant speed of the swimmer

$\begin{array}{l}\mathrm{v} = \sqrt{{\mathrm{v}}_{\mathrm{s}}^2-{\mathrm{v}}_{\mathrm{r}}^2} = \sqrt{(4{)}^2-(3{)}^2}\\ = \sqrt{16-9} = \sqrt{7} \mathrm{m}/\mathrm{s}\\ \tan \mathrm{\theta } = \frac{{\mathrm{v}}_{\mathrm{r}}}{\mathrm{v}} = \frac{3}{\sqrt{7}}\\ \Rightarrow \mathrm{\theta } = {\tan }^{-1}\left(\frac{3}{\sqrt{7}}\right)\text{ of north }\end{array}$

Step 3: Calculate the shortest time taken by the swimmer.

(c) In case (a).
Time is taken by the swimmer to cross the river, $t_1=\frac{d}{v_s}=\frac{d}{4}s$
In case (b).
Time is taken by the swimmer to cross the river
$\begin{array}{l} {\mathrm{t}}_1 = \frac{\mathrm{d}}{\mathrm{v}} = \frac{\mathrm{d}}{\sqrt{7}}\\ \mathrm{As} \frac{\mathrm{d}}{4} < \frac{\mathrm{d}}{\sqrt{7}},\text{ therefore }{\mathrm{t}}_1 < {\mathrm{t}}_2\end{array}$

Hence, the swimmer will cross the river in a shorter time in case (a).