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    I hope it's just 3am talking, but I can't for the life of me work out why BF = 3/2a
    (part iii)

    Have inserted both question and mark scheme

    Please help!!

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    Also, the part about OF and part iv) Thank you to any magician that helps :cry:
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    (Original post by Beenieeee)
    I hope it's just 3am talking, but I can't for the life of me work out why BF = 3/2a
    (part iii)

    Have inserted both question and mark scheme

    Please help!!

    Name:  question.jpg
Views: 90
Size:  55.2 KB
    Attachment 519397519399
    Since F is on the x-axis, the distance BF is the same as the y-coordinate of B (you can see this from the diagram).
    Thus BF = a(1-cos theta) by the given parametric equations
    = a(1-(-1/2)) using theta = 2pi/3
    = a(1+1/2) = 3a/2, as required.
    Similarly, from the diagram, you can see that the distance OF is the same as the x-coordinate of B.
    Thus OF = a(theta - sin theta) by the given parametric equations
    = a(2pi/3 - sqrt(3)/2) using theta=2pi/3.

    For part (iv): Imagine a vertical line going down from C to meet the x-axis. Call G the point at which it meets the x-axis.
    Then we have OE = OF + FG + GE, and from the diagram you can see that FG = BC (they're parallel), so OE = OF + BC + GE. Now by symmetry, OF = GE, so OE = BC + 2OF -> BC = OE - 2OF. (you might be able to just see this from the diagram, but if not, I've just shown you how to derive it).
    From (i), you know that OE = 2pi*a, and from (iii), OF = a(2pi/3 - sqrt(3)/2).
    Thus BC = 2pi*a - 2a(2pi/3 - sqrt(3)/2), and factorising gives BC = a(2pi - 4pi/3 + sqrt(3)) = a(2pi/3 + sqrt(3)).

    Now for AF, consider the triangle AFB. By GCSE-level trigonometry, tan(30) = BF/AF. Thus AF = BF/tan(30). Putting in BF = 3a/2 from (iii) gives AF = (3a/2)/(1/sqrt(3)) = 3a/2 * sqrt(3) = 3/2*sqrt(3)*a.

    For the final bit: we'll need to use point G from earlier again.
    We have that AD = AF + FG + GD. From the diagram, FG = BC, so AD = AF + BC + GD.
    By symmetry, GD = AF, so AD = BC + 2AF. Using previous work, this becomes AD = a(2pi/3 + sqrt(3)) + 2(3/2*sqrt(3)*a) = a(2pi/3 + sqrt(3)) + 3*sqrt(3)*a. Factorising makes this a(2pi/3 + sqrt(3) + 3*sqrt(3))
    = a(2pi/3 + 4*sqrt(3)).
    Hence we set a(2pi/3 + 4*sqrt(3)) = 20, so dividing gives a = 2.22 (3 s.f.).
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    (Original post by Beenieeee)
    I hope it's just 3am talking, but I can't for the life of me work out why BF = 3/2a
    (part iii)

    Have inserted both question and mark scheme

    Please help!!

    Name:  question.jpg
Views: 90
Size:  55.2 KB
    Attachment 519397519399
    Also one more thing: the Examiner's Report for the June 2006 paper (from which this question is taken) states that hardly any candidates made substantial progress on this question, resulting in some extremely low grade boundaries. So don't feel bad if you found it difficult.
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    (Original post by HapaxOromenon2)
    Also one more thing: the Examiner's Report for the June 2006 paper (from which this question is taken) states that hardly any candidates made substantial progress on this question, resulting in some extremely low grade boundaries. So don't feel bad if you found it difficult.
    You're a wizard. Thank you so much! X
 
 
 
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