SOLUTION: A ship at A is to sail to C, 56 km north and 258 km east of A. After sailing N25o10’E for 120 mi to P, the ship is headed toward C. Find the distance of P from C and the requir

Question 1185891: A ship at A is to sail to C, 56 km north and 258 km east of A. After sailing
N25o10’E for 120 mi to P, the ship is headed toward C. Find the distance of P
from C and the required course to reach C.
Answer by CPhill(2189) (Show Source): You can put this solution on YOUR website!
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°

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