Answer:
Step-by-step explanation:
You want the bearing and distance from Charleston, SC, of a boat after it travels at 20 knots from Charleston from 1:30 pm to 4:30 pm due east, then N 18° E until 8:00 pm.
It is helpful if you are familiar with determining hours from clock times, and with the relation between time, speed, and distance. The first leg lasted 3 hours from 1:30 to 4:30. In that time, the boat traveled (20 nmi/h)·(3 h) = 60 nmi. The second leg lasted 3.5 hours from 4:30 to 8:00, so the distance traveled was (20 nmi/h)·(3.5 h) = 70 nmi.
There are several ways you can find the sum of the vectors representing the distance and bearing.
The first attachment shows the solution offered by a geometry app.
The boat is on a bearing of 50.8° from Charleston, at a distance of 105.3 nautical miles.
The second attachment shows the result of using a calculator to find the vector sum. For this, we factored out the speed and used hours for the magnitude of the vectors.
The boat is 105.3 nautical miles on a bearing of 50.8° from Charleston.
You can also find the magnitude of the distance using the law of cosines. The angle between the directions of travel is 90+18 = 108°, so the distance will be ...
 c² = a² +b² -2ab·cos(C)
 c² = 60² +70² -2·60·70·cos(108°) = 11095.74
 c = √11095.74 = 105.3 . . . . nautical miles
The bearing north of east can now be found using the law of sines:
 α = arcsin(sin(108°)·70/105.3) ≈ 39.2°
The bearing clockwise from north is then 90° -39.2° = 50.8°.
60 nmi due east puts the boat at (60, 0) on an x-y plane. Traveling 70 nmi on a bearing 62° counterclockwise from east adds 70(cos(72°), sin(72°)) ≈ (21.63, 66.57) to the coordinates, so the final position is (81.63, 66.57) relative to the origin at Charleston. This is converted to distance and angle by ...
 d = √(x² +y²) = √(81.63² +66.57²) = √11095.74 = 105.3 . . . nautical miles
 α = arctan(66.57/81.63) = 39.2°
The bearing is 90° -α = 50.8°.
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Additional comment
You may notice that our x-y coordinate solution measured the angles counterclockwise from the +x axis, the way angles are conventionally measured on an x-y plane. This requires we subtract the resulting angle from 90° in order to find the bearing.
On the other hand, our calculator solution (attachment 2) used bearing angles directly. If we were to convert these distance∠angle coordinates to rectangular coordinates, they would correspond to (north, east) coordinates, rather than the (east, north) coordinates of an (x, y) plane.
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