Question

A man wants to cross the river to an exactly opposite point on the other bank. If he can row his boat with $\frac{2}{\sqrt{3}}$ times the velocity of the current, then at what angle to the current he must keep the boat pointed?

• A. ${60}^{\circ }$
• B. ${90}^{\circ }$
• C. ${120}^{\circ }$
• D. ${150}^{\circ }$

Answer 1

The correct option is D ${150}^{\circ }$

Let the velocity of boat w.r.t river be $\overrightarrow{V_{bt}}$,


And, let the speed 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 ground must be $0$.

$\Rightarrow \overrightarrow{V_{bt}}\sin \theta -\overrightarrow{V_r}=0$

$\Rightarrow \sin \theta =\frac{\overrightarrow{V_r}}{\overrightarrow{V_{bt}}}$

According to question,

$\overrightarrow{V_{bt}}=\frac{2}{\sqrt{3}}{V}_r$

$\Rightarrow \sin \theta =\frac{\sqrt{3}}{2}$

$\Rightarrow \theta ={60}^{\circ }$

Hence, required angle,

$\varphi ={90}^{\circ }+\theta$

$={90}^{\circ }+{60}^{\circ }$

$={150}^{\circ }$

Hence, option (d) is the correct answer.