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    Just doing one of the latest M3 papers for revision and came across this question which isn't making sense to me (even while looking at the mark scheme). I'm probably missing something obvious but the wording of the question doesn't seem to reflect the diagram drawn in the mark scheme; so would be really grateful if somebody could explain where they've gotten the angles in the question and then how the solution actually works.

    Cheers for the help

    Question Paper 4)b): http://www.edexcel.com/migrationdocu...e_20120614.pdf
    Mark Scheme:
    http://www.edexcel.com/migrationdocu...s_20120816.pdf
    Formulae:
    http://www.edexcel.com/migrationdocu...cal-Tables.pdf
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    (Original post by oli_G)
    ...
    The half angle of the cone (at the vertex V) is \displaystyle\tan^{-1}\frac{a}{2a} and that gives you the 26.6 degrees.

    81.87 arises from using the sum of the angles across the top equals 180. Note they used a more accurate value thatn 26.6 for working out this bit.

    They then take moments about V.
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    (Original post by ghostwalker)
    The half angle of the cone (at the vertex V) is \displaystyle\tan^{-1}\frac{a}{2a} and that gives you the 26.6 degrees.

    81.87 arises from using the sum of the angles across the top equals 180. Note they used a more accurate value thatn 26.6 for working out this bit.

    They then take moments about V.
    Thanks that makes sense now, wasn't in the right frame of mind for mechanics so got confused about what I was doing.

    Cheers
 
 
 
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