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!