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Triangle with median and altitude

  1. Nov 29, 2013 #1

    utkarshakash

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    Gold Member

    1. The problem statement, all variables and given/known data
    In triangle ABC with A as obtuse angle, AD and AE are median and altitude respectively. If BAD = DAE=EAC, then sin^3(A/3)cos(A/3) equals

    2. Relevant equations
    3. The attempt at a solution

    CE = a/2. Let DE = x. Then BD = a/2 - x.
    Let AE = p, AD = q.

    For ΔADB

    [itex]\cos \frac{A}{3} = \dfrac{q^2+c^2-(a/2 -x)^2}{2cq}[/itex]

    I can also write cos A/3 for other two triangles using the same approach but I don't know which one to use in the final expression. Also it will contain p and q which is unknown. I have no idea what sin(A/3) would be. This seems really complicated. :cry:
     
  2. jcsd
  3. Nov 29, 2013 #2

    tiny-tim

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    hi utkarshakash! :smile:
    why use such an awkward triangle? :redface:

    there are three right-angled triangles …

    use them! :wink:
     
  4. Nov 29, 2013 #3

    utkarshakash

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    Gold Member

    How could I miss them?:tongue2: Thanks a lot.
     
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