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This problem involves two ships leaving the same port at the same time but traveling in different directions at different speeds. To solve it, we can apply the law of cosines, as this is essentially a triangle problem where we know two sides and the angle between them.

### Step-by-step solution:

1. **Determine the distance traveled by each ship**:
   - Ship A is traveling at 18 knots for 3 hours. The distance covered by Ship A:
     \[
     d_A = 18 \, \text{knots} \times 3 \, \text{hours} = 54 \, \text{nautical miles}
     \]
   - Ship B is traveling at 24 knots for 3 hours. The distance covered by Ship B:
     \[
     d_B = 24 \, \text{knots} \times 3 \, \text{hours} = 72 \, \text{nautical miles}
     \]

2. **Determine the angle between the two courses**:
   - Ship A is sailing at a bearing of \( 125^\circ \) and Ship B at \( 230^\circ \).
   - The difference in their bearings is:
     \[
     \text{Angle} = 230^\circ - 125^\circ = 105^\circ
     \]
     This is the angle between their paths.

3. **Apply the law of cosines to find the distance between the two ships**:
   The law of cosines states that for a triangle with sides \( a \), \( b \), and \( c \), and the angle \( \theta \) between sides \( a \) and \( b \):
   \[
   c^2 = a^2 + b^2 - 2ab \cos(\theta)
   \]
   Here:
   - \( a = 54 \) (distance of Ship A),
   - \( b = 72 \) (distance of Ship B),
   - \( \theta = 105^\circ \).

   Plugging the values into the formula:
   \[
   d^2 = 54^2 + 72^2 - 2 \times 54 \times 72 \times \cos(105^\circ)
   \]

4. **Calculate the distance**:
   I will now calculate the exact distance.The distance between the two ships after 3 hours is approximately **100.56 nautical miles**.

5. **Find the bearing of Ship B from Ship A**:
   To determine the bearing of Ship B relative to Ship A, we can use trigonometry. Specifically, we use the formula for the angle \( \beta \) between Ship A's direction and the line connecting the two ships:
   \[
   \tan(\beta) = \frac{b \sin(\theta)}{a - b \cos(\theta)}
   \]
   where:
   - \( a = 54 \, \text{nm} \) (distance traveled by Ship A),
   - \( b = 72 \, \text{nm} \) (distance traveled by Ship B),
   - \( \theta = 105^\circ \).

Let's now calculate the bearing of Ship B from Ship A.The bearing of Ship B from Ship A after 3 hours is approximately **168.76°**.

### Final Answer:
- The distance between Ship A and Ship B is **100.56 nautical miles**.
- The bearing of Ship B from Ship A is **168.76°**.

Would you like further details or have any questions?

Here are 5 related questions to expand on this:
1. How would the distance and bearing change if Ship A sailed faster?
2. How would we solve the problem if the angle between the ships' courses was smaller?
3. Can this method be used if the ships followed curved paths instead of straight lines?
4. What if Ship A and Ship B left at different times? How would the calculation change?
5. How does the law of cosines relate to general triangle geometry?

**Tip:** When solving navigation problems, always ensure your bearings are converted into true degrees (0° to 360°) and keep track of directions relative to north.

Two ships leave the same port at the same time. Ship A sails on a course of 125°T at 18 knots while Ship B sails on a course of 230°T at 24 knots. Find, after 3 hours, the distance and bearing of ship B from ship A.

Learn how to calculate the distance and bearing between two ships using trigonometry. This problem uses the law of cosines and bearings to determine the solution after both ships have sailed for 3 hours. Suitable for high school students in grades 11-12.

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