Answer:
(a) 101.77 mi S; 63.59 mi W
(b) S 39.4° W; 131.7 nautical miles
Explanation:
(a)
In the 6 hours from noon until 6 pm, the ship will have traveled a distance of ...
distance = speed × time
distance = (20 nm/h) × (6 h) = 120 nm . . . . . . using "nm" for nautical miles
In (S, W) coordinates, the location of the ship will be ...
120(cos(32°), sin(32°)) ≈ 120(0.848048, 0.529919) ≈ (101.77, 63.59)
The ship will have traveled 101.77 miles south and 63.59 miles west.
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(b)
In the 1 hour from 6 pm to 7 pm, the ship will have traveled an additional 20 miles west. In (S, W) coordinates, its location at 7 pm is (101.77, 83.59). The distance from port will be ...
d = √(101.77² +83.59²) ≈ 131.7 . . . nautical miles
The direction from the port is ...
arctan(83.59/101.77) ≈ 39.4°
The bearing and distance are ...
- S 39.4° W
- 131.7 nautical miles
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Additional comment
The math for bearings and distances works the same for any consistent set of reference directions and angles. Here, all of the points of interest are in the 3rd quadrant of the Cartesian plane. The given angle is measured CW from south. It is convenient to use (south, west) coordinates so that we can use the distance and angle measures directly, without having to translate to (x, y) coordinates.
If bearings are given as angles CW from north, then it is convenient to use (N, E) coordinates.