A

Step 1: Understand the problem setup. We have a ship moving eastward with a velocity of $v_0 = 30 \text{ km/hour}$ along the equator. The wind is blowing from the southeast at an angle $\phi = 60°$ to the equator with a velocity $v = 15 \text{ km/hour}$.

Step 2: Break down the wind velocity into its components. The wind can be decomposed into its eastward and southward components using trigonometric functions.
The eastward component of the wind:
$$ v_{wx} = v \cdot \cos(60°) = 15 \cdot \cos(60°) = 15 \cdot 0.5 = 7.5 \text{ km/hour} $$
The southward component of the wind:
$$ v_{wy} = v \cdot \sin(60°) = 15 \cdot \sin(60°) = 15 \cdot \frac{\sqrt{3}}{2} \approx 12.99 \text{ km/hour} $$

Step 3: Calculate the wind's relative velocity to the ship. The ship has a velocity only in the eastward direction, while the wind has both eastward and southward components. The relative velocity of the wind with respect to the ship in its own reference frame is given by:
$$ v'_{wx} = v_{wx} - v_0 = 7.5 - 30 = -22.5 \text{ km/hour} $$ (west)
$$ v'_{wy} = v_{wy} = 12.99 \text{ km/hour} $$ (south)

Step 4: Now, calculate the magnitude of the relative wind velocity:
$$ v' = \sqrt{(v'_{wx})^2 + (v'_{wy})^2} = \sqrt{(-22.5)^2 + (12.99)^2} $$
$$ = \sqrt{506.25 + 168.68} = \sqrt{674.93} \approx 25.93 \text{ km/hour} $$

Step 5: Next, find the angle $\phi'$ between the equator and the wind direction using tangent:
$$ \tan(\phi') = \frac{v'_{wy}}{-v'_{wx}} = \frac{12.99}{22.5} \quad (The angle is measured counter-clockwise from east towards south) $$
$$ \phi' = \tan^{-1}\left(\frac{12.99}{22.5}\right) \approx 29.61° $$
The angle is therefore in the southwest direction.

Step 6: We report the relative velocity and angle. Thus, the wind velocity relative to the ship is approximately $25.93 \text{ km/hour}$ and the angle $\phi' \approx 29.61°$ from the equator towards the south.

Therefore, the wind velocity relative to the ship is $25.93\text{ km/hour}$, and the angle to the equator is approximately $29.61°$.