how far apart are the two ships?

Question

Two ships leave a harbor at the same time, traveling on courses that have an angle of 100 degrees between them. If the first ship travels at 20 miles per hour and the second ship travels at 25 miles per hour, how far apart are the two ships after 3.1 hours?

Answer 1

It is a triangle ABC with C at the origin

3.1*20 = 62

3.1*25 = 77.5

then a=BC = 62 and b=AC = 77.5

angle BAC = angle C = 100 degrees as given

Law of cosines says:

c^2 = a^2 + b^2 - 2ab * cos 100

= 62^2 + 77.5^2 - 2(62)(77.5)^2 cos 100

= 3844 + 6006.25 - 9610 * cos 100

= 9850.25 - 9610*cos100

=9850.25 - (-1668.76)

= 11519

then , taking the square root of both sides, c = 107.33