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Position Vector

  1. Sep 11, 2013 #1
    1. The problem statement, all variables and given/known data
    The distance travelled by an airplane flying from San Francisco International airport (SFO) to San Jose International (SJC) is 30 nautical miles 36° south of east. Flying from SJC to Tracy, the plane's displacement is 36 nautical miles 52° east of north. What is Tracy's position vector in nautical miles relative to SFO?


    2. Relevant equations

    ##a^2+b^2=c^2##


    3. The attempt at a solution

    I've tried to draw a triangle, but the positions are weird. I know I need to use trig, but I can figure out what formula to use, or where to start even. Someone help!
     
  2. jcsd
  3. Sep 11, 2013 #2

    SteamKing

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    What's weird about the positions? Have you checked your sketch with a map of California? After all, Google Maps has covered the globe.
     
  4. Sep 11, 2013 #3
    Yeah I have, but I don't think I'm getting the angles correctly.

    Here's a photo, SF at the top left, SJ south, Tracy east: http://imgur.com/LSEgciu

    I'm also not entirely sure what sort of trig to use, only just learned what vectors were yesterday.
     
  5. Sep 11, 2013 #4

    SteamKing

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    Well, we can't really provide any suggestions unless we see your work.
     
  6. Sep 14, 2013 #5
    I went to office hours, and I learned that it was just the Law of Cosines.

    such that:

    c^2 = (30^2+36^2-2(30*36)cos(ab).
     
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