Answer
The velocity of the canoe relative to the river is 0.36 m/s at an angle of $51.8^{\circ}$ south of west.
Work Step by Step
Let $E$ be the Earth.
Let $R$ be the river.
Let $C$ be the canoe.
$v_{C/E} = v_{C/R}+v_{R/E}$
$v_{C/R} = v_{C/E} - v_{R/E}$
We can find the east component of $v_{C/R}$
$v_{C/R} = v_{C/E} - v_{R/E}$
$v_{C/R} = (0.40~m/s)~cos(45^{\circ}) - 0.50~m/s$
$v_{C/R} = -0.22~m/s$
Note that the negative sign means that the relative velocity is 0.22 m/s toward the west.
We can find the south component of $v_{C/R}$
$v_{C/R} = v_{C/E} - v_{R/E}$
$v_{C/R} = (0.40~m/s)~sin(45^{\circ}) - 0$
$v_{C/R} = 0.28~m/s$ toward the south
We can find the magnitude of $v_{C/R}$.
$v_{C/R} = \sqrt{(0.22~m/s)^2+(0.28~m/s)^2}$
$v_{C/R} = 0.36~m/s$
We can find the angle south of west.
$tan(\theta) = \frac{0.28~m/s}{0.22~m/s}$
$\theta = tan^{-1}(\frac{0.28}{0.22}) = 51.8^{\circ}$
The velocity of the canoe relative to the river is 0.36 m/s at an angle of $51.8^{\circ}$ south of west.
