Let [tex]\theta[/tex] be the direction the swimmer must swim relative to east. Then her velocity relative to the water is
[tex]\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)[/tex]
The current has velocity vector (relative to the Earth)
[tex]\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath[/tex]
The swimmer's resultant velocity (her velocity relative to the Earth) is then
[tex]\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}[/tex]
[tex]\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath[/tex]
We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:
[tex]\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ[/tex]
which is approximately 41º west of north.
