Two ships leave a harbor at the same time. One ship travels on a bearing Upper S 13 degrees Upper W at 16 miles per hour. The other ship travels on a bearing Upper N 75 degrees Upper E at 11 miles per hour. How far apart will the ships be after 2 ​hours?

To solve this problem, we can treat the paths of the two ships as two sides of a triangle, and we want to find the distance between them after 2 hours. We'll use the Law of Cosines to find this distance, since we know two sides and the angle between them.

### Step-by-Step Breakdown:

1. **Bearings and Speeds:**
   - The first ship is traveling on a bearing \( S13^\circ W \) at 16 miles per hour.
   - The second ship is traveling on a bearing \( N75^\circ E \) at 11 miles per hour.

   The two ships are moving in different directions, and the angle between their directions needs to be calculated.

2. **Calculate the distance traveled by each ship:**
   - In 2 hours, the first ship will have traveled:
     \[
     16 \text{ miles/hour} \times 2 \text{ hours} = 32 \text{ miles}
     \]
   - In 2 hours, the second ship will have traveled:
     \[
     11 \text{ miles/hour} \times 2 \text{ hours} = 22 \text{ miles}
     \]

3. **Determine the angle between the paths:**
   The angle between the two bearings is the sum of the angles measured from north (since they are moving in opposite quadrants). The angle between them is:
   \[
   13^\circ + 75^\circ = 88^\circ
   \]

4. **Apply the Law of Cosines:**
   The Law of Cosines formula is:
   \[
   c^2 = a^2 + b^2 - 2ab \cdot \cos(C)
   \]
   where:
   - \( a = 32 \) miles
   - \( b = 22 \) miles
   - \( C = 88^\circ \)

   Plugging in the values:
   \[
   c^2 = 32^2 + 22^2 - 2 \times 32 \times 22 \times \cos(88^\circ)
   \]

5. **Solve the equation:**
   \[
   c^2 = 1024 + 484 - 2 \times 32 \times 22 \times 0.0349
   \]
   \[
   c^2 = 1024 + 484 - 49.09
   \]
   \[
   c^2 = 1459.91
   \]
   \[
   c \approx \sqrt{1459.91} \approx 38.2 \text{ miles}
   \]

### Final Answer:
After 2 hours, the ships will be approximately **38.2 miles apart**.

---

Would you like further details or clarifications?

Here are five related questions:
1. How would the result change if the speeds of the ships were different?
2. What if the ships traveled for 3 hours instead of 2?
3. How do bearings relate to angles in trigonometry?
4. Could you solve this using vector analysis instead of the Law of Cosines?
5. How does the cosine of an angle near 90 degrees affect the result?

**Tip:** When using bearings in navigation, always remember to account for the angle between directions by adding or subtracting angles based on quadrant locations.

Two ships leave a harbor at the same time. One ship travels on a bearing S 13 degrees W at 16 miles per hour. The other ship travels on a bearing N 75 degrees E at 11 miles per hour. How far apart will the ships be after 2 hours?

Learn how to calculate the distance between two ships after 2 hours using bearings and the Law of Cosines. This problem involves trigonometric concepts and applies to students in Grades 10-12. Detailed step-by-step solution provided.

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