As given that, Speed of river $\left(v_t\right)=3 \mathrm{~m} / \mathrm{s}$ (towards east)
Speed of swimmer $\left(v_s\right)=4 \mathrm{~m} / \mathrm{s}$ (east)
(a) When swimmer starts swimming due north then his resultant velocity
$$
\begin{aligned}
v &=\sqrt{v_r^2+v_s^2}=\sqrt{(3)^2+(4)^2} \\
&=\sqrt{9+16}=\sqrt{25}=5 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$


$$
\begin{aligned}
&\tan \theta=\frac{v_r}{v_s}=\frac{3}{4}=0.75=\tan 36^{\circ} 54^{\prime} \\
&\theta=36^{\circ} 54^{\prime} \text { (with North direction) }
\end{aligned}
$$
Towards west as again perpendicular to axis.
(b) To reach opposite points $B$, let the swimmer should swim at an angle with north direction resultant velocity of the swimmer
$$
v=\sqrt{v_s^2-v_r^2}=\sqrt{(4)^2-(3)^2}
$$
So, $v=\sqrt{16-9}=\sqrt{7} \mathrm{~m} / \mathrm{s}$
$$
\therefore \tan \theta=\frac{v_r}{v}=\frac{3}{\sqrt{7}}
$$
$\theta=\tan ^{-1}\left(\frac{3}{\sqrt{7}}\right)$ of north.

$\tan \theta=1.13=\tan 48^{\circ} 29^{\prime} 30^{\prime \prime}$
$\theta=48^{\circ} 29^{\prime} 30^{\prime \prime}$ from north to west.
(c) Consider width of river $=d_R$ Time taken by the swimmer to cross the river $t_1=\frac{d_R}{v_s}=\frac{d_R}{4}$
When swimmer strike at angle $48^{\circ} 29^{\prime} 30^{\prime \prime}$ its resultant velocity along $\perp$ to river is $v=\sqrt{7} \mathrm{~m} / \mathrm{s}$.
In case (b),
Time taken by the swimmer to cross the river
$$
\begin{aligned}
&t_1=\frac{d}{v}=\frac{d}{\sqrt{7}} \\
&\frac{d}{4} < \frac{d}{\sqrt{7}} \\
&u>\sqrt{7} \text { s. } t_1 < t_2
\end{aligned}
$$
Hence, the swimmer will cross the river in shorter time in case (a).