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Math Help - rectangular

  1. #1
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    rectangular

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

    find  AX
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  2. #2
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    In the triangle \Delta ADY we have \sin (60) = \frac{{AD}}{{AY}}.
    Now solve for AY.
    Why does AY = AX?
    Last edited by Plato; December 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 ABCD, equilateral triangle AXY.
    X is on BC,\;Y is on CD.\;\;AB = 22,\;BC = 14\sqrt{3}

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

    Let a = AX = side of the triangle.

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

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

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

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

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

    . . 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: . \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]: . a \:=\:\frac{14\sqrt{3}}{\cos4.715^o} \:\approx\:24.331

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