Pre-Calculus question! Please Help!

A ship is heading due north at 12 mph. The current is flowing southwest at 4mp. Find the actual bearing and speed of the ship.

please explain on how to arrive at answer.

(in the back of my math textbook it says that the answer is 342.86 degrees and 9.6 mph)

Draw a triangle: AB vertical length 12
Angle B = 45 degrees
BC = 4

b^2 = 4^2 + 12^2 - 2(4)(12)cos(45)
= 92.1177
b = 9.598
b = 9.6

Angle A = 17.14

Bearing = 360 - 17.14 = 342.86

......↑
......|.12 mph due North
......|
......0
..../
../ 4 miles per hour
↓directed into the third quadrant
45° South West

This has a component
in west directon of size

4
▬ = 2√2 mph
√2

and a component in south direction of same magnitude

so it is 2√2

The net North component of the
velocity is

12 - 2√2

and the net west component is

2√2

The actual direction is Northwest {in second quadrant}
and its magnitude satisfies

c² = a² + b²

so
c² = (2√2 )² +(12.-.2√2 )²
....= 8 + 9.1715472875²
= 92.11774901

so c is the square root of this number
and that is
9.59779 which is about 9.6 miles per hour

and since that is the magnitude, it is the hypotenuse
of a triangle with adjacent side 2√2
so
.............2√2
cos Θ = ▬▬=0.29469
..............9.59779

so Θ = inverse cos 0.29469= 72.86°

That angle is North of west. Bearing, for our purposes
is defined as the angle measured from true North CLOCKWISE

so the bearing is 270° to get us from north to west
plus the 72.86° measured to get us back somewhat toward North
but still north west and that is 342.86°

Think in terms of vector components (I'm setting north at being at 90 degrees and southwest at being at 225 degrees; i may be wrong, but the ideas are sound):