document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "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!\r
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Algebra.Com's Answer #225133 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
Break up the ship's travel into x and y components.
\n" ); document.write( "3 miles at 60=(\"+3%2Acos%2860%29\",\"3%2Asin%2860%29\")=(\"1.5\",\"2.598\")
\n" ); document.write( "2 miles at 150=(\"2%2Acos%28150%29\",\"2%2Asin%28150%29\")=(\"-1.732\",\"1.0\")
\n" ); document.write( "At this point the ship's position is
\n" ); document.write( "(1.5-1.732,2.598+1.0)=(-0.232,3.598)
\n" ); document.write( "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
\n" ); document.write( "\"D%5E2=%280.232%29%5E2%2B%283.598%29%5E2\"
\n" ); document.write( "\"D=sqrt%2813%29=3.606\"
\n" ); document.write( "The total distance the ship travels is then,
\n" ); document.write( "\"+Dt=3%2B2%2BD=5%2Bsqrt%2813%29=8.606\" miles
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