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Sail Boat Race Trigonometry, just stuck on first part

  1. Oct 1, 2003 #1
    I've added a picture thats part of the problem.

    A sailboat race course consists of four legs defined by the displacement vectors A, B, C and D shown above.
    The values of the angles are È1 = 420, È2 = 410, and È3 = 270.

    The magnitudes of the first three vectors are A = 3.7 km, B = 5.3 km and C = 4.8 km. The finish line of the course coincides with the starting line.

    The coordinate system for this problem has positive x to the right, positive y as up and counter-clockwise to be a positive angle.

    ------------------------------------------------------------------
    Now first thing I should do is break up each vector into its components, and then add up all the x and y components to get vector D.

    This is what I've tried.

    Vector a
    x-component = 3.7*cos 42=2.75km
    y-component = 3.7*cos42= 2.48km

    vector b
    x component = 5.3*cos 41=4.00km
    y component = 5.3*sin41= 3.48km

    vector c
    x component = 4.8*cos 27=4.27km
    y component= 4.8*sin27= 2.18km
    -------------------------------------------------------------------
    vctor d
    x component = 11.02km
    y component = 8.14km

    So for vector d, just add up the first 3 components in their respective columns. However, when I calculate D's distance => sqrt(11.02^2+8.14^2), I get the incorrect answer of 13.70km.
    So of course, if this answer is wrong, then most likely I figured out one of the components wrong, but I don't know which ones. Please help.
     

    Attached Files:

    Last edited: Oct 1, 2003
  2. jcsd
  3. Oct 1, 2003 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Did you notice that vector B is going back to the LEFT? Its x component is negative. Both the x component and y component of C will be negative. Also notice that once you have added the components of A, B, and C, the result will be the NEGATIVE of D.
     
    Last edited: Oct 2, 2003
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