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Physic Vector Problem

  1. Mar 3, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the components of the vector [tex]\vec{C}[/tex] with length c = 1.00 and angle [tex]\phi = 30.0^{\circ}[/tex] as shown.

    NV_ct.jpg

    2. Relevant equations
    [tex]opp=hypcos(\theta)[/tex]
    [tex]adj=hypcos(\theta)[/tex]


    3. The attempt at a solution

    tried to use the above equations to answer the question, however when i looked in the answers it was incorrect.
    Answer i got was x=0.866 and y=0.5

    When i realized that it was between the vector and the y axis, i found that the angle didn't make contact with the x-axis, so my question is What is the angle that [tex]\vec{C}[/tex] makes with the positive x axis?

    P.S
     
  2. jcsd
  3. Mar 3, 2010 #2

    tiny-tim

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    Hi Paymemoney! :smile:
    erm :redface: … they can't both be right, can they? :biggrin:
    Start at the positive x-axis, go round to the y-axis, then go round some more … that's your angle! :wink:

    But the main thing you've done wrong is that the dotted line marked "x" on your diagram isn't x, is it? :smile:
     
  4. Mar 3, 2010 #3
    oops, meant to be [tex]opp=hypsin(\theta)[/tex]

    ok so i got 600 degrees, what is the next step i should take to solve [tex]\vec{C}[/tex]
     
  5. Mar 3, 2010 #4

    tiny-tim

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    Hi Paymemoney! :smile:

    (have a degree: º and a theta: θ and a phi: φ :wink:)
    hold it!

    however did you get 600º ? :confused:
     
  6. Mar 3, 2010 #5
    well i went around the the axis 2 times o_o
     
  7. Mar 3, 2010 #6

    tiny-tim

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    Still not 600º. :confused:

    Anyway, just go from the x-axis to the y-axis anticlockwise, and then carry on to C.
     
  8. Mar 4, 2010 #7
    ok this is what i have done:

    vectorquess.JPG

    tell me if this is correct.
     
  9. Mar 4, 2010 #8

    tiny-tim

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    Yes, that's exactly correct. :smile:

    The "first quadrant" (top-right) is 0º to 90º, the "second quadrant" (top-left) is 90º to 180º, and so on. :wink:
     
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