SOLUTION: The HMS Pinafore leaves port with bearing S82 degrees E maintaining a speed of 42 miles per hour (that is, with respect to the water). If the ocean current is 3 miles per hour wit

I'm guessing that you rounded off too soon.  You should keep all the 
decimals until the end of the calculation.  Only then should you do
any rounding off.  I will do your problem two ways.

FIRST WAY:

Here are the two vectors involved, the upper shorter black one is the 
vector of the ocean current, the lower black one is the vector of the ship
relative to the ocean. The two blue lines complete the parallelogram, and
the diagonal (the red vector) is the velocity vector of the ship.



The angle between the two given vectors is 180°-(30°+82°) = 68°

Therefore the angle at the bottom right of the parallelogram is
180°-68° = 112°

  

Now we have a case of SAS to solve the lower triangle. The sides
of the 112° angle, are the 40 mph vector of the ship, the short right 
side of the paralallelogram which is 3 units long because it is the 
same length as the ocean vector.  

We use the law of cosines to find the magnitude of the ship's velocity 
vector with respect to the land.









That's the magnitude of the red vector, the ship's velocity vector
with respect to the land.

Now we use the law of sines to find the angle between the the ship's
velocity vector with respect to the ocean (in green), and the ship's
velocity vector with respect to the land.  Suppose that angle is .







So that is how much we must add to the given ships bearing S 82° E relative
to the ocean to get the ship's bearing relative to the land.

So the ship's bearing relative to the land is:

S 82°+3.690551943° E

or

S 85.690551943° E

or to the nearest tenth of a degree:

S 85.7° E

-----------------
SECOND WAY:

You can also do it just by adding components.
We convert the given angles to angles measured counterclockwise
from the "EAST"

short vector's horizontal component = 3cos(60°) = 1.5
long vector's horizontal component = 42cos(352°) = 43.09125889
---------------------------------------------------------------
sum of horizontal components = 43.09125889


short vector's vertical component = 3sin(60°) = 2.598076211
long vector's vertical component = 42sin(352°) = -5.84527024 
------------------------------------------------------------
sum of vertical components = -3.247194029

The resultant is found by the Pythagorean theorem







That is the magnitude of the ship's velocity vector with
respect to the land.  [Same as when done the other way)

The angle is found by getting the inverse tangent of
the quotient of the two component sums:





So the ships bearing east of south is found by

90° - 4.309448057° = 85.69055194°

or

S 85.690551943° E

or to the nearest tenth of a degree:

S 85.7° E

Edwin