A man wants to cross the river to an exactly opposite point on the other bank. If he can row his boat with twice the velocity of the current, then at what angle to the current he must keep the boat pointed?

The correct option is C ${120}^{\circ }$

Let the velocity of boat w.r.t river be $\overrightarrow{V_{br}}$
Let the velocity of the river be $\overrightarrow{V_r}$
In order to cross the river with the shortest path, the net horizontal component of velocity of boat w.r.t river must be zero.
⟹$|\overrightarrow{V_{br}}\sin \theta |=|\overrightarrow{V_r}|$
$\sin \theta =\frac{|\overrightarrow{V_r}|}{|\overrightarrow{V_{br}}|}$ ;
According to question $|\overrightarrow{V_{br}}=2|\overrightarrow{V_r}|$
$\Rightarrow \sin \theta =\frac{1}{2}$
$\Rightarrow \theta ={30}^{\circ }$
Hence, required angle is
${90}^{\circ }+{30}^{\circ }={120}^{\circ }$