Pre-Calculus Question?

I have a test tomorrow and I can't seem to find answer for this problem.
Help! Please show me the steps.

A cruise ship maintains an average speed of 15 knots in going from San Juan, Puerto Rico, to Barbados, West Indies, a distance of 600 nautical miles. To avoid a tropical storm, the Captain heads out of San Juan in a direction of 20 degress off a direct heading to Barbados. The Captain maintains the 15 knot speed for 10 hours, after which time the path to Barbados becomes clear of storms. Through what angle should the Captain turn to head directly to Barbados?

Draw a triangle from San Juan (point A) to
Barbados (point B) to the point where the
captain turns the ship after clearing the
storm (point C).

We know that angle CAB is 20 degrees.
Segment AB is 600.
Segment AC is 15*10 = 150.

We can determine the length of segment BC by
using the law of cosines:

(BC)^2 = (AC)^2 + (BA)^2 - 2(AC)(BA) cos(angle BAC)
BC^2 = 150^2 + 600^2 - 2*150*600*cos(20 degrees)
BC^2 = 22500 + 360000 - 180000 * 0.939692621
BC^2 = 213355.328220000
BC = 461.904

Now we can determine angle BCA via the law of sines:

sin(BCA)/600 = sin(20)/461.904
sin(BCA) = (sin(20)/461.904) * 600
sin(BCA) = 0.342020143 / 461.904 * 600
sin(BCA) = 0.44427432063805465800
BCA = arcsin(0.44427432063805465800)
BCA = 26.376919355 or 153.623080645

From the diagram, we can see that we want the
obtuse angle, so BCA = 153.623080645

So from the current course (which is 20 degrees
off of a direct line from San Juan to Barbados),
the captain will need to turn:

course change = 180 - angle BCA
course change = 180 - 153.623080645
course change = 26.376919355

Answer:

Turn 26.4 degrees back toward Barbados.

Incidentally, the remaining leg of the trip is 461.9
nautical miles.

.

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