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  1. #1
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    rectangular

    $\displaystyle ABCD$ rectangular , $\displaystyle AXY$ Equilateral Triangle
    X∈BC , Y∈CD and $\displaystyle AB=22 , BC=14\sqrt 3$

    find$\displaystyle AX $
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  2. #2
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    In the triangle $\displaystyle \Delta ADY$ we have $\displaystyle \sin (60) = \frac{{AD}}{{AY}}$.
    Now solve for AY.
    Why does AY = AX?
    Last edited by Plato; Dec 6th 2007 at 10:19 AM.
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  3. #3
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    Hello, perash!

    I hope I interpreted this correctly.
    Even then, I have an intricate solution.
    . . Maybe someone can simplify it.


    Rectangle $\displaystyle ABCD$, equilateral triangle $\displaystyle AXY.$
    $\displaystyle X$ is on $\displaystyle BC,\;Y$ is on $\displaystyle CD.\;\;AB = 22,\;BC = 14\sqrt{3}$

    Find $\displaystyle AX.$
    Code:
                X
        B * - -*- - - - - - - * C
          |        *          |
          |   * 60   * a     |
          |              *    |
          |  *a             * |
       22 |               60 * Y
          | *             *   |
          |           * a     |
          |* 60  *           |
          |   * θ             |
        A * - - - - - - - - - * D
                  14√3

    Let $\displaystyle a = AX$ = side of the triangle.

    Let $\displaystyle \theta \,=\,\angle YAD$
    Then $\displaystyle 30^o-\theta \,=\,\angle XAB$

    In right triangle $\displaystyle YBA\!:\;\;\cos\theta \:=\:\frac{14\sqrt{3}}{a}\quad\Rightarrow\quad a \:=\:\frac{14\sqrt{3}}{\cos\theta} $ .[1]

    In right triangle $\displaystyle XBA\!:\;\;\cos(30-\theta) \:=\:\frac{22}{a}\quad\Rightarrow\quad a \:=\:\frac{22}{\cos(30-\theta)} $ .[2]

    Equate [1] and [2]: .$\displaystyle \frac{14\sqrt{3}}{\cos\theta} \:=\:\frac{22}{\cos(30-\theta)}\quad\Rightarrow\quad14\sqrt{3}\cos(30-\theta) \:=\:22\cos\theta$

    . . $\displaystyle 14\sqrt{3}(\cos30\cos\theta + \sin30\sin\theta) \:=\:22\cos\theta$ . $\displaystyle \Rightarrow\quad 14\sqrt{3}\left(\frac{\sqrt{3}}{2}\cos\theta + \frac{1}{2}\sin\theta\right) \:=\:22\cos\theta$

    . . $\displaystyle 21\cos\theta + 7\sqrt{3}\sin\theta \:=\:232\cos\theta \quad\Rightarrow\quad 7\sqrt{3}\sin\theta \:=\:\cos\theta \quad\Rightarrow\quad \frac{\sin\theta}{\cos\theta}\:=\:\frac{1}{7\sqrt{ 3}}$

    Hence: .$\displaystyle \tan\theta \:=\:\frac{1}{7\sqrt{3}}\quad\Rightarrow\quad\thet a \:=\:\tan^{\text{-}1}\!\left(\frac{1}{7\sqrt{3}}\right) \:\approx\:4.715^o$

    Substitute into [1]: .$\displaystyle a \:=\:\frac{14\sqrt{3}}{\cos4.715^o} \:\approx\:24.331$

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