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Angle Between Vectors given Ratio of Resultants, why is this wrong?

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data

    Vectors A and B have the same magnitude. Given that the magnitude of A + B is 75 times greater than the magnitude of A - B, find the angle between them?


    2. Relevant equations

    We know that A=B, so:

    2AB+2ABCos[itex]\theta[/itex]=75(2AB-2ABCos[itex]\theta[/itex])

    Given that A=B

    2A2+2A2Cos[itex]\theta[/itex]=75(2A2-2A2Cos[itex]\theta[/itex])


    3. Attempt at a solution

    2A2+2A2Cos[itex]\theta[/itex]=75(2A2-2A2Cos[itex]\theta[/itex])

    Couldn't get latex to work for the fraction so:

    2A2+2A2Cos[itex]\theta[/itex] divided by (2A2-2A2Cos[itex]\theta[/itex])=75

    Factor out 2A2:

    [itex]\frac{1+Cos}{1-cos}=75
    1+Cos[itex]\theta[/itex]=75-75[itex]\theta[/itex]
    -74=-76Cos[itex]\theta[/itex]
    Cos[itex]\theta[/itex]=74/76
    [itex]\theta[/itex]=Cos-1(74/76)
    [itex]\theta[/itex]=13.2 deg

    Apparently this is wrong, how so?
     
  2. jcsd
  3. Sep 26, 2011 #2
    Hey,
    Actually, the way I see it, verifying your logic and procedure, it seems fine!
    Are you sure they're not asking for the solution in radians?
    It might also help, to get abit of order in your calculations, to mark it as this:
    Assuming the angle between the two is theta(like you said), but the ratio, for the sake of generality, we'll call alpha, that gives us:
    [itex]
    \large
    |A+B|=\alpha |A-B|, |A|=|B| \Longrightarrow A^2+2A^2\cos{\theta}+A^2 = \alpha(A^2-2A^2\cos{\theta}+A^2)
    [/itex]
    Dividing by 2*A^2:
    [itex]
    \large
    1+\cos{\theta} = \alpha(1-\cos{\theta}) \Longrightarrow \cos{\theta} = \frac{\alpha-1}{\alpha+1}
    [/itex]
    Check to see that you have the proper answer key.
    Are there any other considerations you may have overlooked?
    Daniel
     
  4. Sep 26, 2011 #3
    Thanks for that, noted for next time. This is the question in exact form, I can't find anything I'm missing.

    DH7iN.png
     
  5. Sep 26, 2011 #4
    Try entering it in radians, namely: 0.229 ~ 0.23.
    What's the typical accuracy, digit significance asked? are two decimal places usually sufficient?
    If so, try either 0.229, 0.23.
    Daniel
     
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