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**Hint**To find the angle between the direction of swimming of the man and that of the bank, first find the angle between the direction of swimming and the vertical. Find the component of the swimming of the man to the vertical to find the time taken to cross the river.

Formula used: In this solution we will be using the following formulae;

\[v = \dfrac{d}{t}\] where \[v\] is the velocity of a body in a particular direction, \[d\] is the displacement traversed by the body in a direction, and \[t\] is the time elapsed for the displacement to be covered.

\[hy{p^2} = op{p^2} + ad{j^2}\] where \[hyp\] signifies hypotenuse side of a right angled triangle, \[opp\] is opposite side, and \[adj\] is the adjacent side.

**Complete Step-by-Step solution:**

To find the angle between the targeted velocity of the swimming of the man (3km/h) and that of the velocity of the river, we must calculate the angle \[\theta \] first. This can be calculated as in

\[\tan \theta = \dfrac{2}{3}\]

\[ \Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{2}{3}} \right) = 41.7^\circ \]

Now, since, the river flow is parallel to the river bank, then the angle the man should try to swim

\[\alpha = 90 + \theta = 131.7^\circ \]

To find the time taken to cross the river, we must find the actual velocity of the man. We can use Pythagoras theorem, which says

\[hy{p^2} = op{p^2} + ad{j^2}\] where \[hyp\] signifies hypotenuse side of a right angled triangle, \[opp\] is opposite side, and \[adj\] is the adjacent side.

Hence,

\[{x^2} = {3^2} - {2^2} = 9 - 4 = 5\]

\[ \Rightarrow x = \sqrt 5 km/h\]

Now, from the equation

\[v = \dfrac{d}{t}\] where \[v\] is the velocity of a body in a particular direction, \[d\] is the displacement traversed by the body in a direction, and \[t\] is the time elapsed for the displacement to be covered, we have

\[\sqrt 5 = \dfrac{{0.5}}{t}\] (since 500 m is \[0.5km\])

Hence,

\[t = \dfrac{{0.5}}{{\sqrt 5 }} = 0.22hr\] or \[13.4\] minutes.

**Note:**Alternatively, one could calculate the real velocity of the man by finding the component of the 3km/h on the vertical axis, as in

\[v = 3\cos 41.7^\circ = 2.23\]

This is equivalent to \[\sqrt 5 \]

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