For the swimmer to go straight the sin component of his velocity should cancel the river velocity.
So, we have [tex]v_{1} = v_2sin[/tex]θ
This angle θ is the angle shall the swimmer point upstream from the shore.
0.32 = 0.85 sin[/tex]θ
θ = [tex]22.12^0[/tex]
The velocity of swimmer across the river is given by cos componet of his velocity
v = 0.85 cosθ = 0.85 cos 22 = 0.79 m/s