First thing to do is sketch the figure. In geometry it is important that you have the correct figure or diagram....especially with bearings.

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.