The resultant velocity is given as the resultant vector magnitude as:

`V_R = sqrt(4.9^2 + 10.7^2) = 11.8 m/s`

The direction of this vector is given as

`theta = tan^(-1) (4.9/10.7)` = **24.6 degrees** (in South-East direction or 24.6 degrees towards south from east direction)

Since the counter-clockwise direction is positive,...

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The resultant velocity is given as the resultant vector magnitude as:

`V_R = sqrt(4.9^2 + 10.7^2) = 11.8 m/s`

The direction of this vector is given as

`theta = tan^(-1) (4.9/10.7)` = **24.6 degrees** (in South-East direction or 24.6 degrees towards south from east direction)

Since the counter-clockwise direction is positive, resultant is **-24.6 degrees** from the original heading.

The time taken to cross the river can be calculated as the ratio of distance to the velocity (component of resultant velocity in that direction)

thus, `t = (distance)/V_e = 1373/(V_R cos theta)`

= **128.4 sec**

Hope this helps