The figure would be a triangle ABC where
AB = 650 mi
BC = 810 mi
angle ABC = angle B = 48 +36 = 84 degrees
Why 48 +36?
Because at point B, if you drop a vertical line to signify the South axis, then the bearing of BA is S 48deg W ....just the opposite of the bearing of AB, which is given as N 48deg E.
So in the figure, there are two known sides and their included angle. So use the Law of Cosines to find the 3rd side.
(CA)^2 = (650)^2 +(810)^2 -2(650)(810)cos(84deg)
CA = 984.14 mi --------answer.
Now you can get the angle C by the Law of Sines:
650/ sinC = 984.14 /sin(84deg)
angle C = arcsin[650*sin(84) /984.14deg)
angle C = 41.06 deg, or, say, 41 degrees.
Then, for the bearing of CA,
here is one way,
draw a North-South crossline at point C.
it is easy to see now the bearing of CA.
it is N (36 +41 = 77) W....or N 77deg W -------answer.