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ABC is an equilateral triangle

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data
    Based on my understanding of the question the diagram should be like this

    Trigo.jpg


    ABC is an equilateral triangle with side 2a.Two circle are are drawn , with one of them crossing the points A,B and C.another circle crossess B and C with A as centre

    Prove that the area enclosed by the two minor arcs BC are

    (2a^2 / 9) x (3[itex]\sqrt{}3[/itex] - ∏)




    2. Relevant equations

    Area of Segment of an Triangle
    A= 1/2 a^2 (θ-sinθ)

    3. The attempt at a solution

    Minor Arc Segment = Outer Segment - Inner Segment

    Outer Segment

    A=1/2 a^2 x (120 - sin120)
    = (120-[itex]\sqrt{}3[/itex] / 2) a^2

    Inner Segment

    A=1/2 2a^2 x (60-sin60)
    =(120-[itex]\sqrt{}3[/itex])a^2


    Note:Update the working Steps in tmr ,sleep now! is 3am !
     
    Last edited: Nov 5, 2011
  2. jcsd
  3. Nov 5, 2011 #2

    SteamKing

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    Re: Trigo!!

    I think you are supposed to use radians to measure the angles.
     
  4. Nov 5, 2011 #3

    HallsofIvy

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    Re: Trigo!!

    Yes. It doesn't matter for the "[itex]sin(\theta)[/itex]" of course, but it does for the first [itex]\theta[/itex] in "[itex]\theta- sin(\theta)[/itex]".
     
  5. Nov 5, 2011 #4
    Re: Trigo!!

    hmm thanks i think i got it
     
  6. Nov 5, 2011 #5
    Re: Trigo!!

    Suppose ?? Can I Continue to Asking In The Same Thread ?
    By Using (Equation Reducible to Quadratic Form)

    3-3cosX = 2sin^2 X
    2sin^2X - 3cosX-3

    As (sin^2 X) =1-cos2X / 2
    2(1-cos2X/2) = 1-cos2X

    1-1-cos2X -3cosX - 3
    cos2X - 3cosX -3

    As cos2X = 2 cos^2 X-1
    So

    2cos^2 X - 1 -3cosX - 3

    Let cos X be y

    2y^2 -1 - 3y - 3 = 0
    2y^2 - 3y - 4 = 0
    ( ) ( ) = 0

    Y1 = 2.35
    Y2 = -0.85

    No Real Number LOL!

    Or Must Solving Using

    Equation for the form a sin θ + - b cos θ = c
    R cos A = θ--------------1
    R sin A = θ --------------2
     
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