Question 1185891
Here's how to solve this navigation problem:

**1. Convert the initial bearing and distance to coordinates:**

*   The ship sails N25°10'E for 120 nautical miles (we'll assume the problem uses nautical miles).
*   Convert the bearing to a decimal degree: 25° + (10'/60) = 25.167°
*   Calculate the change in latitude (north/south): 120 * cos(25.167°) ≈ 108.64 NM
*   Calculate the change in longitude (east/west): 120 * sin(25.167°) ≈ 50.84 NM

**2. Determine the coordinates of point P:**

*   Point A is our origin (0, 0).
*   Point P's coordinates are (50.84, 108.64) relative to A.

**3. Determine the coordinates of point C relative to A:**

*   Point C is 258 km east and 56 km north of A.  We need to convert these to nautical miles.  1 nautical mile is approximately 1.852 km.
*   East: 258 km / 1.852 km/NM ≈ 139.31 NM
*   North: 56 km / 1.852 km/NM ≈ 30.24 NM
*   Point C's coordinates are (139.31, 30.24) relative to A.

**4. Determine the coordinates of point C relative to point P:**

*   Subtract the coordinates of P from the coordinates of C:
    *   East: 139.31 - 50.84 ≈ 88.47 NM
    *   North: 30.24 - 108.64 ≈ -78.4 NM

**5. Calculate the distance from P to C:**

*   Use the Pythagorean theorem: distance = sqrt(east² + north²)
*   distance = sqrt(88.47² + (-78.4)²) ≈ sqrt(7827.84 + 6146.56) ≈ sqrt(13974.4) ≈ 118.21 NM

**6. Calculate the bearing from P to C:**

*   Use the arctangent function to find the angle: angle = arctan(east / north)
*   angle = arctan(88.47 / -78.4) ≈ -48.42°

*Since the change in North is negative and change in East is positive, the bearing will be in the fourth quadrant.*

*   To convert to a standard bearing, we measure clockwise from North. Since we are in the fourth quadrant, we will use 360 - 48.42 to get the bearing.

*   Bearing = 360 - 48.42 ≈ 311.58 or N 48.42° W

**Answers:**

*   Distance from P to C: Approximately 118.21 nautical miles.
*   Course to reach C from P: Approximately N 48.42° W or 311.58°