2 ships leave a port at the same time. When ship A is 180 miles due east of the port, Ship B is 170 miles from

Question

the port. and 140 miles from ship A.

What is ship B's barring.

Answer 1

"Bearing" always confuses me. I think you want the angles. You have a triangle ABC, where C is the port; A & B are the ships; and a, b, and c are the sides opposite A, B, and C.

We have a = 170, b = 180, and c = 140. We want to get angles A, B, and C.

To do that, we use the Law of Cosines.

a^2 = b^2 + c^2 - 2bc cos A
2bc cos A = b^2 + c^2 - a^2
cos A = (b^2 + c^2 - a^2) / (2bc)

cos A = (180^2 + 140^2 - 170^2) / (2 x 180 x 140)
cos A = (18^2 + 14^2 - 17^2) / (2 x 18 x 14)
cos A = 0.45833
A = 62 degrees, 43 minutes (answer)

Similarly, cos B = (17^2 + 14^2 - 18^2) / (2 x 17 x 14)
cos B = 0.33824
B = 70 degrees, 14 minutes (answer)

cos C = (17^2 + 18^2 - 14^2) / (2 x 17 x 18)
cos C = 0.68137
C = 47 degrees, 3 minutes (answer)

Check: angles A + B + C = 180 degrees (correct)

You figure out which angle you want for your bearing. My guess is you want C.