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Math Help - Special Right Triangles Help Needed

  1. #1
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    Special Right Triangles Help Needed

    Well I am new here but I am really stuck on one of my problems that I was assigned to for HW.

    It is..

    The length of an altitude of an equilateral triangle is 12 feet. Find the length of a side of the triangle.

    I kinda get the concept but this one just blows my mind.
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  2. #2
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    Re: Special Right Triangles Help Needed

    Try solving 12^2 = (2x)^2+x^2
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  3. #3
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    Re: Special Right Triangles Help Needed

    Hello, GeometryBeginner!

    Welcome aboard!


    The length of an altitude of an equilateral triangle is 12 feet.
    Find the length of a side of the triangle.

    Did you make a sketch?

    Code:
                  *
                 /|\
                / | \
             x /  |  \ x
              /   |   \
             /    |12  \
            /     |     \
           /      |      \
          * - - - + - - - *
          :  x/2  :  x/2  :
    See the right triangle?

    Pythagorus says: . \left(\tfrac{x}{2}\right)^2 + 12^2 \:=\:x^2

    Now solve for x.

    You should get: . x \,=\,8\sqrt{3}
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Special Right Triangles Help Needed

    A trigonometric approach:

    \sin(60^{\circ})=\frac{12}{x}

    x=\frac{12}{\sin(60^{\circ})}=\frac{12}{\frac{ \sqrt{3}}{2}}=\frac{24}{\sqrt{3}}=8\sqrt{3}
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