The man should swim in a direction of 60 degrees North of East.

To determine the optimal direction, we can use the concept of vector addition. Let's assume that the width of the river is 100m. In still water, the man could swim directly towards the opposite bank and cover the distance of 100m in 10 minutes. However, due to the current of the river, he would be displaced downstream by 5m/min, as shown in the following diagram:

```
RIVER FLOW
|
----|----> EAST
|
|
|
MAN'S PATH
```

To minimize the time to reach the opposite bank, the man should aim to end up at a point directly across from his starting point, taking into account the drift caused by the river flow. One way to achieve this is to break down the velocity of the man's swimming into two components: one perpendicular to the river flow and one parallel to it. The perpendicular component will allow him to cross the river, while the parallel component will cancel out the effect of the river flow.

Let's call the direction of the perpendicular component θ (measured from the North direction) and the magnitude of the parallel component v. Then we can write:

cos(θ) = v/10 (since the man's swimming speed in still water is 10m/min)
sin(θ) = 5/v (since the drift caused by the river flow is 5m/min)

Solving these equations simultaneously, we get:

v = 10sin(θ)/√(sin^2(θ)+25)
cos(θ) = sin(θ)/√(sin^2(θ)+25)

To minimize the time to cross the river, we want to minimize the total distance the man has to swim, which is given by:

d = 100/sin(θ)

Substituting the expression for v from above, we get:

d = 100√(sin^2(θ)+25)/sin(θ)

To find the minimum value of d, we can take the derivative with respect to θ and set it to zero:

d' = 100(25cos(θ)-sin(θ)√(sin^2(θ)+25))/sin^2(θ)^(3/2) = 0

Solving for θ, we get:

tan(θ) = 5/√(25-4) = 5/3

Therefore, θ = 60 degrees North of East. This means that the man should aim to swim at an angle of 60 degrees North of East, as shown in the following diagram:

```
RIVER FLOW
|
----|----> EAST
|
\ |
\ |
\θ | 100m
\ |
\ |
\ |
\ |
\ |
\|
MAN
```

In this direction, the perpendicular component of the man's swimming velocity will be 10cos(60) = 5m/min, which will allow him to cross the river in 20 minutes (since the width of the river is 100m). The parallel component of his velocity will be 10sin(60) = 5√3 m/min, which will cancel out the effect of