you should, however, learn how to use the law of cosines to solve the problem, even though it's not necessary in this case.
you also make use of parallel lines and alternate interior angles in this problem.
the law of cosines formula is c^2 = a^2 + b^2 - 2ab * cos(C).
a,b,c are sides of the triangle.
C is an angle of the triangle that is between sides a and b.
each side of the triangle is opposite it's associated angle, which has the same name, only in capital letters, rather than small letters.
it helps to draw a sketch of your triangle and to label it in such a way so as to conform to the requirements of the law of cosines.
my sketch is shown below:
it is NOT drawn to scale, which can be a problem in visualizing what the triangle looks like.
i have tried to draw it somewhat to scale, so it's kind of like drawn to scale, but not exactly.
south 42 degrees west means the angle is drawn to the left of the vertical axis, pointing in the south west direction.
south 48 degrees east means the angle is drawn to the right of the vertical axis, pointing in the south east direction.
in the diagram, the triangle formed is ABC.
side c is opposite angle C.
side a is opposite angle A.
side b is opposikte angle B.
the line segment DG is a vertical axis to point C.
the line segment FE is a vertical is a vertical axis to point A.
angle GCB is the 48 degrees angle south east of line segment CG with a vertex at point C.
angle EAC is the 42 degree angle south west of line segment FE with a vertex at point A.
since vertical line segments EF and GD are parallel to each other, then angles EAC and ACD are alternate interior angles of parallel lines and therefore angle ACD is equal to 42 degrees.
since angles GCB and ACB and ACD form a straight line, then the sum of these 3 angles must be equal to 180 degrees.
this makeks angle ACB equal to 180 - 42 - 48 = 90 degrees.
that's the serendipity here, which makes use of the law of cosines not necessary, but we'll use it just so you know how to.
we have the lengths of sides a and b and we have the angle between them.
side a is equal to 27 miles.
side b is equal to 80 miles.
angle C is equal to 90 degrees.
using the law of cosines, we can find the measure of side c.
the formula is c^2 = a^2 + b^2 - 2ab * cos(C).
that becomes c^2 = 80^2 + 27^2 - 2 * 80 * 27 * cos(90).
solve for c^2 to get c^2 = 7129.
that makes c = square root of 7129.
that's your answer becuse the port is point A and the final destination is point B.
the distance between them is the length of the line segment AB which is equal to c which is equal to sqrt(7129) which is equal to 84.43340571 miles which is equal to 84 miles rounded to the nearest mile.
now the serendipity.
since angle ACB is a 90 degree angle, than triangle ABC is a right triangle.
this makes c the hypotenuse and a and b legs of a right triangle.
using pytharogus theorem, we get c^2 = a^2 + b^2 which makes c^2 equal to 80^2 + 27^2 which makes c^2 equal to 7129 which makes c equal to square root of (7129), etc.
bottom line:
your triangle formed is triangle ABC.
if it was not a right triangle, then you would have needed to use the law of cosines.
since it was a right triangle, you didn't need to, and could get what you needed by using the pythagorus theorem.
as i indicated earlier, labeling the triangle so that it fits the generic formula is helpful, if you have the luxury of being able to do so.
otherwise, you need to translate the designations of the formula to the designations of your triangle.
i did not try to find angles B or A because they weren't necessary to find in order to solve the problem.
since this is a right triangle, they can be found using the trigonometric functions of a right triangle.
if it wasn't a right triangle, they could be found using the law of sines or the law of cosines.