C

Step 1: Define the velocities.
The man's swimming velocity vector can be broken down into components:
- North Component (V_{mn}): $V_{mn} = 5 \cos(30°)$
- East Component (V_{me}): $V_{me} = 5 \sin(30°)$
Thus, we calculate the components:
$V_{mn} = 5 \cdot \frac{\sqrt{3}}{2} = 4.33$ km/h (North)
$V_{me} = 5 \cdot \frac{1}{2} = 2.5$ km/h (East)

Step 2: The cyclist's velocity is purely along the East direction at 10 km/h.
Step 3: Now we need to find the relative velocity of the man with respect to the cyclist.
- Cyclist's Velocity (East): 10 km/h
- Man's Velocity:
East Component: 2.5 km/h
North Component: 4.33 km/h

Step 4: Calculate the relative velocity:
Relative velocity in East (V_{relE}) = V_{me} - 10 = 2.5 - 10 = -7.5 km/h (West)
Relative velocity in North (V_{relN}) = 4.33 km/h (North)

Step 5: Now, we can find the magnitude of the relative velocity vector:
\( V_{rel} = \sqrt{(-7.5)^2 + (4.33)^2} = \sqrt{56.25 + 18.75} = \sqrt{75} = 8.66 \text{ km/h} \)

Step 6: Find the angle of this velocity relative to the North direction using the tangent function:
\( \theta = \tan^{-1}\left(\frac{4.33}{7.5}\right) \approx 30.4° \text{ North of West} \)

Therefore, the man appears to swim at a speed of 8.66 km/h, approximately 30.4° North of West.