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    Im currently on part b.
    What does the question mean by saying q < 180?
    Also, I plugged in the y/x values into the equations getting:

    \cos (100p - q)^{\circ} = 0 for A(100, 0) and
    \cos (220p - q)^{\circ} = 0 for B(220, 0).

    \therefore 100p - q = 90 and  220p - q = 90

    When I try solving I get something like p = 0, which is rubbish. What am I doing incorrectly?
    Thanks

    PS just for the reference for part a, is the best approach to change the interval from 0 < x < 360 to 0 < 2x < 360, find the solutions, add 30 then divide by 2?
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    (Original post by UKBrah)
    Im currently on part b.
    What does the question mean by saying q < 180?
    Also, I plugged in the y/x values into the equations getting:

    \cos (100p - q)^{\circ} = 0 for A(100, 0) and
    \cos (220p - q)^{\circ} = 0 for B(220, 0).

    \therefore 100p - q = 90 and  200p - q = 90

    When I try solving I get something like p = 0, which is rubbish. What am I doing incorrectly?
    Thanks

    PS just for the reference for part a, is the best approach to change the interval from 0 < x < 360 to 0 < 2x < 360, find the solutions, add 30 then divide by 2?
    Well, A is your "primary" solution, so B must be a secondary solution.
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    Didnt clock that mate, sorry
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    (Original post by UKBrah)
    Didnt clock that mate, sorry
    If I have

    \cos x = y

    x=\cos^{-1} y

    AND

    x=360 - cos^{-1} y

    make sense?
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    (Original post by Indeterminate)
    If I have

    \cos x = y

    x=\cos^{-1} y

    AND

    x=360 - cos^{-1} y

    make sense?
    So x = 90, 270?..

    How does that exactly help with my sim equations?
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    P must be 0 you can see that this must be true by just looking at the simultaneous equations brah
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    \cos(100p-q) = 0

    \cos(220p-q) = 0

    You can work out the period of the function from this, normally roots of a cosine function would be \pi = 180 ^\circ apart, here however they are only 120 ^\circ apart. So:

    p = \frac{180}{120} = \frac{3}{2}

    Now, the q is easily determined by noticing it's just going to translate the normal cosine function left/right (parallel to the x-axis) so work out where the roots would normally lie with the function \cos(\frac{3}{2}x), and you can work out the q from noting the coordinates of A, B and C.
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    (Original post by Noble.)
    \cos(100p-q) = 0

    \cos(220p-q) = 0

    You can work out the period of the function from this, normally roots of a cosine function would be \pi = 180 ^\circ apart, here however they are only 120 ^\circ apart. So:

    p = \frac{180}{120} = \frac{3}{2}

    Now, the q is easily determined by noticing it's just going to translate the normal cosine function left/right (parallel to the x-axis) so work out the roots would normally lie with the function \cos(\frac{3}{2}x), and you can work out the q from noting the coordinates of A, B and C.
    +1.

    That didn't appeal to me at first
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    (Original post by raiden95)
    P must be 0 you can see that this must be true by just looking at the simultaneous equations brah
    p does not equal 0, it wouldn't be a cosine function if p = 0.
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    (Original post by Noble.)
    p does not equal 0, it wouldn't be a cosine function if p = 0.
    Oh I was looking at the simultaneous equation
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    (Original post by raiden95)
    Oh I was looking at the simultaneous equation
    Well, if p is zero then y = cos(-q) and since q is a constant y is just a straight line.
 
 
 
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