SOLUTION: A ship is traveling at a abearing of 60 degrees for 3 miles then turns and travels at a bearaing of 150 degres for 2 miles then returns to where it started. Find the distance trav

Algebra ->  Trigonometry-basics -> SOLUTION: A ship is traveling at a abearing of 60 degrees for 3 miles then turns and travels at a bearaing of 150 degres for 2 miles then returns to where it started. Find the distance trav      Log On


   



Question 314751: A ship is traveling at a abearing of 60 degrees for 3 miles then turns and travels at a bearaing of 150 degres for 2 miles then returns to where it started. Find the distance traveled by the ship.
I am having a hard time even starting this problem. I tried drawing a 30 degree angle (using "North" as the terminal side and the 3 miles along the side that I drew going northeast). I thought I would use the Law of Sines or Cosines but I can't even begin the problem!
Thank you!

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Break up the ship's travel into x and y components.
3 miles at 60=(+3%2Acos%2860%29,3%2Asin%2860%29)=(1.5,2.598)
2 miles at 150=(2%2Acos%28150%29,2%2Asin%28150%29)=(-1.732,1.0)
At this point the ship's position is
(1.5-1.732,2.598+1.0)=(-0.232,3.598)
In order to return back the ship will have to travel 0.232 in the x, and -3.598 in the y. The total distance is
D%5E2=%280.232%29%5E2%2B%283.598%29%5E2
D=sqrt%2813%29=3.606
The total distance the ship travels is then,
+Dt=3%2B2%2BD=5%2Bsqrt%2813%29=8.606 miles