PLEASE HELP: POP QUIZ: Two ships leave a port at 9 A.M. One travels at a bearing of N 53 ◦ W at 12 miles per hour and the other travels at a bearing of S 67 ◦ W at 16 miles per hour. Approximate how far apart they are at noon that day.
Answer in units of mi PLEASE HELP: POP QUIZ: Two ships leave a port at 9 A.M. One travels at a bearing of N 53 ◦ W at 12 miles per hour and the other travels at a bearing of S 67 ◦ W at 16 miles per hour. Approximate how far apart they are at noon that day.
Answer in units of mi @Mathematics

Question

PLEASE HELP: POP QUIZ: Two ships leave a port at 9 A.M. One travels at a bearing of N 53 ◦ W at 12 miles per hour and the other travels at a bearing of S 67 ◦ W at 16 miles per hour. Approximate how far apart they are at noon that day.
Answer in units of mi  PLEASE HELP: POP QUIZ: Two ships leave a port at 9 A.M. One travels at a bearing of N 53 ◦ W at 12 miles per hour and the other travels at a bearing of S 67 ◦ W at 16 miles per hour. Approximate how far apart they are at noon that day.
Answer in units of mi  @Mathematics

hero · Answer

Didn't I help you with this already?

anonymous · Answer

nope.

mertsj · Answer

After you draw your picture, you see that you are talking about a triangle with a 60 degree angle.  The sides that form the 60 degree angle are 3(12) and 3(16) or 36 and 48.  So now you can use the law of cosines using the 60 degree angle as C and the required distance as c.  So:  c^2 = 48^2+36^2 - 2(48)(36)cos 60
           c^2 = 2304+1296-3456(.5)
           c^2 = 3600-1728
           c^2 = 1872
           c = 43.3 miles