A ship leaves port on a course with a bearing of 20 degrees. After sailing in that direction for 76 kilometers, it changes course to a bearing of 105 degrees and travels for 143 km. If the ship went directly back to Port, what bearing did it take?

The triangle has a side of 76 pointed "up" - 20 degrees right of the y-axis (bearing) rather than off of the x-axis (angle). The turn to a bearing of 105 is a slope downward (greater than 90 from the y-axis, or under the x-axis). That makes a side of 143. The question is what "bearing" must be taken to reach the origin again. From the dimensions, the outward-bound angle after the turn is 95. Drawing out the diagram should help. Proceed from there trigonometrically. A quick geometric construction shows a return bearing of 285 (360 - 75). Distance approximately 168.

To find the bearing the ship took to return directly to port, we can use trigonometry. We'll first calculate the displacement vector from the final position of the ship to the port, and then find the angle or bearing of that vector with respect to north.

1. Calculate the displacement vector from the final position to the port: a. From the final position, move 76 km in the direction opposite to the last leg of the journey (105 degrees from north). b. Then move 143 km in the direction opposite to the initial leg of the journey (20 degrees from north).

2. Use trigonometry to find the angle or bearing of the displacement vector with respect to north.

3. To find the bearing, subtract the angle from the direction of north.

Using trigonometry:

a. ( x = 76 \times \cos(105^\circ) + 143 \times \cos(20^\circ) ) b. ( y = 76 \times \sin(105^\circ) + 143 \times \sin(20^\circ) )

Calculate ( \theta = \arctan\left(\frac{y}{x}\right) ) in degrees.

The bearing would then be ( 90^\circ - \theta ) or ( 270^\circ + \theta ), depending on which quadrant the ship's displacement vector lies in.

Finally, convert the bearing to the appropriate format (0° to 360°).