Velocity of an oblique projectile in its flight

Resultant of the velocity of man in still water

${\text{v}}_{\text{m}}\left(=\text{OA}\right)$  and the velocity of water is $\left({\text{v}}_{\omega }\right)\left(=\text{OB}\right)$  along$\text{OP}\text{and}\left(\text{v}\right)\left(=\text{OC}\right)$

${\Delta }^{ie}\text{OCA}\sin x=\text{AC/OA}$

$=\text{v}\omega \text{/vm}=\text{5/10}$

$\sin x=1/2$

$\sin x=\sin 30°\Rightarrow x=30°$

$\therefore$ The man should start swimming at an angle of $30°$  east of west