SOLUTION: A ship leaves a port and sails for 4 hours on a course of 78° at 18 knots. Then the ship changes its course to 168° and sails for 6 hours at 16 knots. After the 10 hours (a) what

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Question 1198575: A ship leaves a port and sails for 4 hours on a course of 78° at 18 knots. Then the ship changes its course to 168° and sails for 6 hours at 16 knots. After the 10 hours (a) what is the distance of the ship from the port and(b) what is the bearing from the port to the ship?
Answer by proyaop(69) (Show Source):
You can put this solution on YOUR website!
**1. Calculate Distances**
* **Leg 1:**
* Speed = 18 knots
* Time = 4 hours
* Distance = Speed * Time = 18 knots * 4 hours = 72 nautical miles
* **Leg 2:**
* Speed = 16 knots
* Time = 6 hours
* Distance = Speed * Time = 16 knots * 6 hours = 96 nautical miles
**2. Determine Coordinates (using Law of Cosines)**
* **Let:**
* A: Starting point (port)
* B: Position after Leg 1
* C: Final position of the ship
* **Calculate the angle between legs (∠ABC):**
* ∠ABC = 168° - 78° = 90°
* **Use the Law of Cosines to find the distance from the port (AC):**
* AC² = AB² + BC² - 2 * AB * BC * cos(∠ABC)
* AC² = 72² + 96² - 2 * 72 * 96 * cos(90°)
* AC² = 5184 + 9216 - 0
* AC² = 14400
* AC = √14400 = 120 nautical miles
**3. Determine the Bearing (using Law of Sines)**
* **Let:**
* ∠BAC be the angle between AB and AC
* **Use the Law of Sines:**
* sin(∠BAC) / BC = sin(∠ABC) / AC
* sin(∠BAC) / 96 = sin(90°) / 120
* sin(∠BAC) = 96 / 120
* sin(∠BAC) = 0.8
* ∠BAC = arcsin(0.8) ≈ 53.13°
* **Calculate the Bearing:**
* Bearing from the port to the ship = Initial course (78°) + ∠BAC
* Bearing = 78° + 53.13° = 131.13°
**Therefore:**
* **(a) Distance from the port:** 120 nautical miles
* **(b) Bearing from the port to the ship:** 131.13°
**Note:**
* A nautical mile is approximately 1.15 miles.
* Bearings are typically measured clockwise from North (0°).
* This solution assumes a perfectly flat Earth and does not account for factors like tides, currents, or wind, which can affect a ship's actual position.