Question

A man wishes to swim across a river $0.5{\rm{ km}}$ wide. If he can swim at the rate of $2{\rm{ }}{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {\rm{h}}}} \right.} {\rm{h}}}$ in still water and the river flows at the rate of $1{\rm{ }}{{{\rm{km}}} {\left/{\vphantom {{{\rm{km}}} {\rm{h}}}} \right.} {\rm{h}}}$. The angle (w.r.t the flow of the river) along which he should swim so as to reach a point exactly opposite his starting point, will be
(1) $60^\circ$
(2) $120^\circ$
(3) $145^\circ$
(4) $90^\circ$

Answer 1

We will represent the given situation graphically to understand it clearly. We will be using the basic trigonometric functions to find the angle made by the man with the flow of water when he is swimming in the opposite direction.

Complete step by step answer:
Given:
The width of the river is $b = 0.5{\rm{ km}}$.
The speed of flow of water in the river is ${V_r} = 1{\rm{ }}{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}$.
The speed with which man can swim is ${V_m} = 2{\rm{ }}{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}$.
We can assume that man is swimming at an angle $\theta$ with the vertical in the opposite direction and the angle made by the man with respect to the flow of the river is $\phi$.
We can consider that the river is flowing from left to right and man is in its opposite direction. Therefore we can represent it graphically as below:
For triangle ABC, using trigonometric function sine, we can write:
$\sin \theta = \dfrac{{{V_r}}}{{{V_m}}}$
We will substitute $1{\rm{ }}{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}$ for ${V_r}$ and $2{\rm{ }}{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}$ for ${V_m}$ in the above expression to find the value of the angle made by the man with the vertical.

$$\sin \theta = \dfrac{{1{\rm{ }}{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}}}{{2{\rm{ }}{{{\rm{km}}} {\left/ {\vphantom {{{\rm{km}}} {\rm{h}}}} \right. } {\rm{h}}}}}\\\ \Rightarrow\sin \theta = \dfrac{1}{2}$$

Taking the inverse of sine on both sides of the above equation, we get:

$$\theta = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)\\\ \Rightarrow\theta = 30^\circ$$

We will write the expression for the angle made by the man with respect to the flow of water, which is equal to the summation of the right angle and angle made by the man with vertical.
$\phi = \theta + 90^\circ$
We will substitute $30^\circ$ for $\theta$ in the above expression to get the value of $\phi$.

$$\phi = 30^\circ + 90^\circ \\\ \therefore\phi = 120^\circ$$

Therefore, the angle made by the man with the flow of water is $120^\circ$, and option (2) is correct.

Note: : Additional method: We can resolve the velocity of man into vertical and horizontal components. Using the equation of equilibrium for x-direction, we can find the value of angle $\theta$ and using geometry; we can find the value of the angle $\phi$.