Two swimmers 'A'&'B', one located on one side and other on the another side of a river are situated at a distance 'D' from each other. Line joining them is making an angle theta with the direction perpendicular to flow. The speed of each swimmer with respect to still water is 'u' & speed of river flow is v_r .

Let one of the swimmer (say 1) crosses the river along AB, which is obviously the shortest path. Time taken to cross the river by the swimmer 1.
t1​=v′2−v02​​d​, (where AB=d is the width of the river) (1)
For the other swimmer (say 2), which follows the quickest-path, the time taken to cross the river.
t2​=v′d​ (2)
In the time t2​, drifting of the swimmer 2, becomes
x=v0​t2​=v′v0​​d, (using equation 2) (3)
If t3​ be the time for swimmer 2 to walk the distance x to come from C to B (shown in figure above), then 
t3​=ux​=v′uv0​d​ (using equation 3) (4)
According to the problem t1​=t2​+t3​
or, v′2−v02​​d​=v′d​+v′uv0​d​
On solving, we get
u=(v′21−v02​​)−21​−1v0​​=