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Angles of a triangle

  1. May 14, 2013 #1
    1. The problem statement, all variables and given/known data
    In a ##\Delta ABC##, angle A is greater than angle B. If the measures of A and B satisfy the equation ##3\sin x-4\sin^3 x-0.75=0##, then angle C is equal to
    A)##\pi/3##
    B)##\pi/2##
    C)##2\pi/3##
    D)##5\pi/6##

    2. Relevant equations



    3. The attempt at a solution
    The given equation can be rewritten as ##12\sin x-16\sin^3 x-3=0##. This is a cubic equation in ##\sin x##, how am I supposed to solve this? :confused:

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. May 14, 2013 #2

    SteamKing

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    Let y = sin x, then you have a cubic in y.

    You could also try manipulating one of the trig identities for sine of the multiple of an angle.
     
  4. May 14, 2013 #3

    NascentOxygen

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    Staff: Mentor

  5. May 14, 2013 #4
    Sorry but I am still lost, how am I supposed to solve ##16y^3-12y+3=0##? :confused:
     
  6. May 14, 2013 #5
  7. May 14, 2013 #6
    Cool! :cool:

    I totally forgot about that. :redface:

    That gives ##\sin 3x=3/4##, hence
    [tex]x=\frac{1}{3}\arcsin\frac{3}{4}, \frac{1}{3}\left(\pi-\arcsin\frac{3}{4}\right)[/tex]

    ##C=\pi-(A+B)## where ##A+B=\pi/3##, therefore ##C=2\pi/3##.

    Thank you David! :smile:

    No, I would never use the alternative method, that isn't in my syllabus and you already gave such a nice method. :tongue2:
     
  8. May 14, 2013 #7

    Curious3141

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    It's interesting that the 0.75 doesn't matter a whit here in solving it.
     
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