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Another Vector Question

  1. Sep 17, 2007 #1
    1. The problem statement, all variables and given/known data
    http://www.tunerspec.ca/school/question7.jpg

    3. The attempt at a solution
    Not sure where to start. I want to get the x & y coordinates of rAB but I'm not sure how.
     
  2. jcsd
  3. Sep 17, 2007 #2

    learningphysics

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    Get the angle of vector OB from the positive x - axis going counterclockwise...
     
  4. Sep 17, 2007 #3
    How? :uhh:
     
  5. Sep 17, 2007 #4

    learningphysics

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    Can you get the angle of OA from the positive x-axis?
     
  6. Sep 17, 2007 #5
    Using

    [tex]tan x = \frac{711j}{418i}[/tex]

    [tex]x = 59.55^{\circ}[/tex]
     
  7. Sep 17, 2007 #6

    learningphysics

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    good. what's the relationship between that OA and angle... and the OB angle?
     
  8. Sep 17, 2007 #7
    I'm not sure...

    If this ends up being easy I'm going to feel stupid. :tongue:
     
  9. Sep 17, 2007 #8
    Label the angle which is 59.55*. You have a triangle and 3 sides are known, use the cosine law to determine any angle right? From that you can get the angle that [tex]|r_{oa}+r_{ab}|[/tex] makes with the origin.

    Then you can use the angle and sine/cosine to find the coords of the point B or you can set up a unit vector and multiply by the magnitude of the [tex]|r_{oa}+r_{ab}|[/tex] to get the components.
     
    Last edited: Sep 18, 2007
  10. Sep 18, 2007 #9

    learningphysics

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    Oops... I thought I posted a reply. Sorry about that!

    Use the angle AOB, and subtract it away from the angle of OA... that gives the angle of OB...
     
  11. Sep 18, 2007 #10
    I'm getting a length of 824.77m for side OA. It looks too big according to the diagram, is it right?

    Assuming it is...

    I used the cosine law to determine the angle for AOB (43.99) and subtracted that from angle x (determined previously.) This gave me an angle of 15.56 from the x-axis to OB. Then using the length of 606m and the angle of 15.56 I determined the x and y coordinates to be 583.79m and 162.56m, respectively.

    Is that right?
     
  12. Sep 18, 2007 #11

    learningphysics

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    yup. looks right to me.
     
  13. Sep 18, 2007 #12
    Thank you for the help. :)
     
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