A

Step 1: Define the vectors involved.
The velocity of the boat relative to the earth (V_BE) is 10 km h^-1 at 37º north of east.
Convert this to its component form:
V_BE_x = 10 * cos(37º) and V_BE_y = 10 * sin(37º).

Step 2: Calculate the components:
V_BE_x = 10 * 0.7986 ≈ 7.986 km h^-1 (eastward)
V_BE_y = 10 * 0.6018 ≈ 6.018 km h^-1 (northward)

Step 3: Note the contributions from wind and current:
Wind velocity (V_W) = -2 km h^-1 (westward, negative in x)
Current velocity (V_C) = -4 km h^-1 (southward, negative in y)

Step 4: Write the resulting velocity of the boat due to its own power (V_BP):
V_BP = V_BE - V_W - V_C
V_BP_x = V_BE_x + 2 km h^-1 (eastward)
V_BP_y = V_BE_y + 4 km h^-1 (northward)

Step 5: Substitute values:
V_BP_x = 7.986 + 2 = 9.986 km h^-1 (eastward)
V_BP_y = 6.018 + 4 = 10.018 km h^-1 (northward)

Step 6: Find the magnitude:
\|V_BP\| = \sqrt{(9.986)^2 + (10.018)^2} \approx \sqrt{99.72 + 100.36} = \sqrt{200.08} \approx 14.14 km h^-1.

Step 7: Find the direction:
\theta = \tan^{-1}\left(\frac{10.018}{9.986}\right) \approx \tan^{-1}(1.003) \approx 45.0º north of east.

Therefore, the magnitude is approximately 14.14 km h^-1 and the direction is approximately 45º north of east.