Maths Help

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A,B,C,D,E,F are 6 consecutive points on the circumference of a circle such that AB=BC=CD=10,DE=EF=FA=22. If the radius of the circle is √n, what is the value of n?

  • Maths Help -

    We have 3 isosceles triangles with a base of 10 and
    3 isosceles triangles with a base of 22
    let the central angle of each of the smaller be x and the angle of each of the larger be y
    3x + 3y = 360°
    x+y = 120°

    for the smaller using the cosine law
    10^2 = √n^2 + √n^2 - 2√n √n cosx
    100 = n+n-2ncosx
    2ncosx = 2n-100
    cosx = (n-50)/n

    similarly for the larger triangle:
    2ncosy = 2n- 484
    cosy = (n-242)/n

    cos(x+y) = cosx cosy - sinx siny

    so we need sinx and sin y

    Make a sketch of a right -angled triangle with base
    n-50, hypotenuse = n and height of h
    By Pythagoras:
    h^2 + (n-50)^2 = n^2
    h^2 = n^2 - (n-50)^2 = 100n - 2500
    h = √(100n-2500) = 10√(n-25)
    sinx = 10√(n-25)/n

    in the same way:
    h2 = 22√(n-121)
    siny = 22√(n-121)/n

    cos(x+y) = cosx cosy - sinx siny
    cos(120°) = (n-50)/n * (n-242)/n - (10√(n-25))/n * 22√(n-121))/n)
    -1/2 = (n-50)/n * (n-242)/n - (10√(n-25))/n * 22√(n-121))/n)

    What a horrible equation:
    So I relied on Wolfram to do all the drudgery and amazingly got

    n = 268

    http://www.wolframalpha.com/input/?i=%28n-50%29%2Fn+*+%28n-242%29%2Fn+-+%2810√%28n-25%29%29%2Fn+*+22√%28n-121%29%29%2Fn%29+%3D+-1%2F2

    Check:
    cosx = (n-50)/n = 218/268
    x = 35.567°
    cosy = (n-121)/n = 26/268
    y = 84.433°
    x+y = 35.3567 + 84.433 = 120° !!!!!! YEahhh

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