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Thread: Equilateral last

  1. #1
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    Equilateral last

    ABC is an equilateral triangle.
    |AE|=|EC|=3 units
    |DC|=2|BD|
    CF is perpendicular to DE

    Find |AF|=X=?

    Equilateral last-kasa.png
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  2. #2
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    Re: Equilateral last

    Show some work?
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  3. #3
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    Re: Equilateral last

    My work is a mess.
    AE=3
    BD=2
    DC=4
    I drew a parallel to AB from D. I tried congruent triangles but couldn't find anything. I need a clue.
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  4. #4
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    Re: Equilateral last

    I would start with trigonometry.

    Use Law of Cosines:
    $|ED|^2 = 3^2+4^2-2\cdot 3\cdot 4 \cos(60^\circ) = 13$
    So, $|ED| = \sqrt{13}$

    Next, use Law of Sines:
    $\dfrac{\sin 60^\circ}{\sqrt{13}} = \dfrac{\sin \left(\angle DEC\right)}{4}$

    $\angle DEC = \arcsin\left(\dfrac{2 \sqrt{39}}{13}\right)$

    That gives
    $\angle ECF = 90^\circ-\arcsin\left(\dfrac{2 \sqrt{39}}{13}\right)$
    $\angle AFC = 120^\circ - \angle CEF = 30^\circ+\arcsin\left(\dfrac{2 \sqrt{39}}{13}\right)$

    Next, use Law of Sines again:

    $\dfrac{\sin \left(\angle ECF\right)}{x} = \dfrac{\sin\left(\angle AFC \right)}{6}$
    $\dfrac{\sqrt{13}}{13x} = \dfrac{\dfrac{1}{2}\dfrac{\sqrt{13}}{13}+\dfrac{ \sqrt{3} }{2}\dfrac{2\sqrt{39}}{13}}{6}$
    $\dfrac{\cancel{\sqrt{13}}}{\cancel{13}x} = \dfrac{7\cancel{\sqrt{13}}}{2\cdot 6\cdot \cancel{13}}$

    $x = \dfrac{12}{7}$
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  5. #5
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    Re: Equilateral last

    Many Thanks for your help. But I must solve it with the classic method. I can't use trigonometry like you did. For example I have never used arcsin function in a problem unless I was asked to do so.
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  6. #6
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    Re: Equilateral last

    I found the solution without trigonometry.
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