Set up a vector diagram to calculate the velocity of the canoe relative to the river.
To do this use the relative motion formula.
Put into words, the velocity of a with respect to c is equal to the velocity of a with respect to b plus the velocity of b with respect to b.
You will find:
Where we dont know `V_(canoe,river)`
We have 2 lengths of the triangle, and the angle between those 2 lengths. For simplicity, lets now call the lengths we do know b and c and the angle between them A.
The length opposite the angle A is what we require as this is the magnitude of the velocity of the canoe (namely, its speed) relative to the river.
We can calculate the length a using the Cosine Rule:
a^2 = b^2 + c^2 - (2bc cosA)
a = sqrt (b^2 + c^2 - 2bc cos A)
Now, we have that the lengths b and c are
b = 0.460 m/s
c = 0.600 m/s
and that the angle A = 45 degrees = pi/4 radians
Finally then we have
a = sqrt ( 0.46^2 + 0.6^2 - (0.46)(0.6)cos (pi/4))
= sqrt ( 0.5716 - 0.276/sqrt(2))
= sqrt (0.3764) = 0.6135 m/s
The magnitude of the canoe relative to the river is 0.614 m/s. However this is westwards on the East/West axis, so that the canoe is drifting southwards and in the opposite direction to the river on the East/West axis relative to the river. You can tell this from your vector diagram. The precise angle of relative drift can be calculated with the Sine Rule or Cosine Rule.