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The Soc for People of 'GRDCT2008' Mk VI watch

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    Ahaha Cs!! oh **** you capitals
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    (Original post by GHOSH-5)
    Yes. I came so much, the closet overflowed :indiff:
    ...I see what you did there....
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    This thread just made a turn for the worse, and I wasn't even involved.

    :rolleyes:
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    (Original post by ArchedEdge)
    This thread just made a turn for the worse, and I wasn't even involved.

    :rolleyes:
    Even worse now you're here :hmmm:
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    (Original post by GHOSH-5)
    Even worse now you're here :hmmm:
    To think, you may be practically neighbours in September :moon:
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    (Original post by natty_d)
    To think, you may be practically neighbours in September :moon:
    Not quite. He'll be a 5 minute walk down the road, and I might not be there (******* STEP).
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    (Original post by GHOSH-5)
    Not quite. He'll be a 5 minute walk down the road, and I might not be there (******* STEP).
    I have every faith you'll do well, Ghosh :console:

    But I did put 'may be' in there :p:
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    (Original post by natty_d)
    I have every faith you'll do well, Ghosh :console:

    But I did put 'may be' in there :p:


    Fair point. :p:
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    (Original post by GHOSH-5)
    Even worse now you're here :hmmm:
    Hai bbz

    :awesome:
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    Ghosh help me bro.

    Stuck on C, how you find the area of the parallelogram.

    And then how to approach D.

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    (Original post by ArchedEdge)
    Ghosh help me bro.

    Stuck on C, how you find the area of the parallelogram.

    And then how to approach D.

    c) Consider the first quadrant: Let the tangent to the ellipse in the first quadrant be called L. It has equation  5y\sin \alpha + 4x\cos \alpha = 20 . Let Y be the point where L meets the y-axis and X be the point where L meets the x-axis. You can find the coordinates of X and Y quite easily. Hence what is the area of triangle XYO (O is the origin)? Hence what is the area of the whole parallelogram?
    Spoiler:
    Show
     X = (5/ \cos \alpha , 0) and  Y = (0, 4/\sin \alpha ).
    Spoiler:
    Show
    \mathrm{Area \ of \ XYO} = \dfrac{10}{\sin \alpha \cos \alpha} = \dfrac{20}{2\sin \alpha \cos \alpha} = \dfrac{20}{\sin 2\alpha} .
    Spoiler:
    Show
     \mathrm{Area \ of \ parallelogram} = 4 \times \mathrm{Area \ of \ XYO} , due to symmetry.
    d) The area of the two types of wood are equal iff half of the area of the parallelogram = the area in the parallelogram but surrounding the ellipse.

    Hence we need to solve:

     \dfrac{1}{2} \left( \mathrm{Area \ of \ parallelogram} \right) = \dfrac{80}{\sin 2\alpha} - 20\pi .
    Spoiler:
    Show
    I think that  \alpha = \frac{1}{2} \sin^{-1} \left( \frac{2}{\pi} \right) , but I haven't got a pen and paper at hand to check.


    edit: cheers Natty :love:
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    (Original post by GHOSH-5)
    c) Consider the first quadrant: Let the tangent to the ellipse in the first quadrant be called L. It has equation  5y\sin \alpha + 4x\cos \alpha = 20 . Let Y be the point where L meets the y-axis and X be the point where L meets the x-axis. You can find the coordinates of X and Y quite easily. Hence what is the area of triangle XYO (O is the origin)? Hence what is the area of the whole parallelogram?
    Spoiler:
    Show
     X = (5/ \cos \alpha , 0) and  Y = (0, 4/\sin \alpha ).
    Spoiler:
    Show
    \mathrm{Area \ of \ XYO} = \dfrac{10}{\sin \alpha \cos \alpha} = \dfrac{20}{2\sin \alpha \cos \alpha} = \dfrac{20}{\sin 2\alpha} .
    Spoiler:
    Show
     \mathrm{Area \ of \ parallelogram} = 4 \times \mathrm{Area \ of \ XYO} , due to symmetry.
    d) The area of the two types of wood are equal iff half of the area of the parallelogram = the area in the parallelogram but surrounding the ellipse.

    Hence we need to solve:

     \dfrac{1}{2} \left( \mathrm{Area \ of \ parallelogram} \right) = \dfrac{80}{\sin 2\alpha} - 20\pi .
    Spoiler:
    Show
    I think that  \alpha = \frac{1}{2} \sin^{-1} \left( \frac{2}{\pi} \right) , but I haven't got a pen and paper at hand to check.


    edit: cheers Natty :love:
    Ahhh yes I forgot about the equation thing they gave you earlier.


    And I just got a lil confused about the wording of the last question
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    On hold at Manchester Med school, what a bunch of gays.
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    How rude of them!
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    I feel for you :flute:
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    (Original post by v1oXx-)
    On hold at Manchester Med school, what a bunch of gays.
    Sorry to hear that but, you're better off at Barts tbh. Don't come to Manchester!
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    I need to have a complete sort out of all my folders
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    BBB from Warwick...:ninja:
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    (Original post by Champagne Supernova)
    BBB from Warwick...:ninja:
    win! congratsss!




    for anyone that might know; RE: the ucas deadline thing that was extended to 22nd Jan, does this mean i could still add a choice on track for equal consideration? or is it only for schools that had snow issues and specifically requested a deadline extension (which im not sure if ours did )? infact, does it even apply for peoples forms that have already been processed?
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    (Original post by John Locke)
    win! congratsss!




    for anyone that might know; RE: the ucas deadline thing that was extended to 22nd Jan, does this mean i could still add a choice on track for equal consideration? or is it only for schools that had snow issues and specifically requested a deadline extension (which im not sure if ours did )? infact, does it even apply for peoples forms that have already been processed?




    I think it's applicable for all scenarios. :s: The UCAS site doesn't give much info, and if the UCAS/Apps forum can't provide an answer, then phone UCAS? :sigh:
 
 
 
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