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Part 1
A ship leaves port on a bearing of 44.0degrees north of east and travels 12.9 mi. The ship then turns due east and travels 4.8 mi. How far is the ship from​ port, and what is its bearing from​ port?
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Part 1
The ship is 
  
enter your response here miles from the port.
​(Do not round until the final answer. Then round to the nearest tenth as​ needed.)

To solve this, we can break the problem into two parts using vector components and trigonometry.

### Step 1: Break down the ship's path into components.

#### First leg of the journey:
- The ship travels 12.9 miles at a bearing of 44° north of east.
  - The horizontal component (x-direction, east) is \( 12.9 \times \cos(44^\circ) \).
  - The vertical component (y-direction, north) is \( 12.9 \times \sin(44^\circ) \).

#### Second leg of the journey:
- The ship then turns due east and travels 4.8 miles.
  - Since the ship is traveling due east, this leg only affects the horizontal (x-direction) by adding 4.8 miles. There is no vertical (north-south) change in position.

### Step 2: Calculate total displacement.
- The total horizontal displacement (x) is:
  \[
  x_{\text{total}} = 12.9 \times \cos(44^\circ) + 4.8
  \]
- The total vertical displacement (y) is:
  \[
  y_{\text{total}} = 12.9 \times \sin(44^\circ)
  \]

### Step 3: Use the Pythagorean theorem to find the total distance from the port.
\[
\text{Distance from the port} = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2}
\]

### Step 4: Find the bearing.
The bearing is the angle measured clockwise from north. Using trigonometry:
\[
\theta = \tan^{-1}\left(\frac{x_{\text{total}}}{y_{\text{total}}}\right)
\]
Since the direction is east of north, this angle needs to be converted into a bearing by subtracting it from 90°.

Let's calculate each part.

### Calculations:
1. \( x_{\text{first leg}} = 12.9 \times \cos(44^\circ) \approx 9.277 \) miles
2. \( y_{\text{first leg}} = 12.9 \times \sin(44^\circ) \approx 8.937 \) miles
3. Total horizontal displacement: \( x_{\text{total}} = 9.277 + 4.8 = 14.077 \) miles
4. Total vertical displacement: \( y_{\text{total}} = 8.937 \) miles

Using the Pythagorean theorem to find the distance from port:
\[
\text{Distance} = \sqrt{(14.077)^2 + (8.937)^2} \approx \sqrt{198.162 + 79.879} \approx \sqrt{278.041} \approx 16.68 \text{ miles}
\]

### Bearing Calculation:
\[
\theta = \tan^{-1}\left(\frac{14.077}{8.937}\right) \approx \tan^{-1}(1.575) \approx 57.4^\circ
\]
Since this is measured clockwise from north, the bearing is \( 90^\circ - 57.4^\circ = 32.6^\circ \) east of north.

### Final Answer:
- The ship is approximately **16.7 miles** from the port.
- The bearing from the port is approximately **32.6° east of north**.

Would you like further details or clarifications on this? Here are some related questions:

1. How do bearings differ from angles in navigation?
2. What is the importance of vector components in navigation?
3. How can the Pythagorean theorem be applied in other real-life situations?
4. Why is it necessary to use trigonometric functions when solving bearing problems?
5. What role does the inverse tangent function play in determining bearings?

**Tip:** Always double-check the angles and directions used in navigation problems to ensure you’re measuring from the correct reference point!

A ship leaves port on a bearing of 44.0 degrees north of east and travels 12.9 mi. The ship then turns due east and travels 4.8 mi. How far is the ship from port, and what is its bearing from port?

Learn how to determine the distance and bearing of a ship from port using trigonometric concepts like vector components and the Pythagorean theorem. This problem involves calculating a ship's path as it travels 12.9 miles at a 44° bearing and then 4.8 miles due east, resulting in a distance of approximately 16.7 miles from port and a bearing of about 32.6° east of north. Suitable for students in Grades 10-12.

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