A ship travels 59km on a bearing of 39 degrees, and then travels on a bearing of 129 degrees for 151km. Find the distance from the starting point to the end of the trip, to the nearest kilometer.

Here is how I worked out the problem, but when I plug everything into the calculator it gives me a figure that is not an option. After reviewing what I did, could someone please tell me if I'm missing something.

X coordinate of the final point: 59cos(39degrees) + 151 cos (129degrees)= -49.2
Y coordinate of the final point: 59 sin (39 degrees) + 151 sin (129degrees)= 154.5

d=square root (-49.2)^2+ (154.5)^2= 162.14 then I squared that answer and came up with 12.73 but this is not one of the answer choices.
Choices are A.) 37 km B.) 46km C.)210km D.)162km

I guess I'm confused if I should continue to square root or just leave it as 162. Thanks a bunch!

In marine navigation, directions are calculated from the North (000°) going towards the right (East = 090°, South = 180° and do on)

For a problem like that (where longitudes are not required) we would use the components "northings" and "eastings"
northings = number of kilometres towards the north (negative if towards the south)
eastings = numbers of km towards the east (negative if to the west)

Given a distance and a direction (the correct word should be "course" not bearing - but that is not important for the problem):

northings = distance * cosine(course)
eastings = distance * sin(course)

Total northings = 59*cos(39°) + 151*cos(129°) = -49.1758 km
Total eastings = 59*sin(39°) + 151*sin(129°) = +154.4789 km

The final position will be south and east of the initial position.

final distance = √( northings^2 + eastings^2) = 162.1172 km
final direction = atan(eastings/northings) = S 72.34° E = 180° - 72.34° = 107.66°

When you found the 162 km, you had already done the square root. No need to do it again.

The correct description of what you did should have been:
d=square root (-49.2)^2+ (154.5)^2= (162.14)^2
showing that the 162.14 is the square root of the sum (because the sum is the square).