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Math Help - Finding side "x" (triangles)

  1. #1
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    Finding side "x" (triangles)

    Can't figure this one out.

    The answer must be in exact value.


    *Picture has been fixed to add the missing side length
    Attached Thumbnails Attached Thumbnails Finding side "x" (triangles)-.bmp  
    Last edited by shawli; April 4th 2010 at 05:17 AM.
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    Quote Originally Posted by shawli View Post
    Can't figure this one out.

    The answer must be in exact value.
    Dear shawli,

    I think to solve this problem the lengh of at least one of the sides must be given. Are you sure the length of any side is not given??
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    Quote Originally Posted by shawli View Post
    Can't figure this one out.

    The answer must be in exact value.
    There's not enough information. You need to give at least one known length.
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    Sorry! I missed a label, but I've reposted the picture with the correction.
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    Quote Originally Posted by shawli View Post
    Sorry! I missed a label, but I've reposted the picture with the correction.
    Dear shawli,

    If the other side of the rectangle is y;

    tan45^0=\frac{x}{y}

    tan60^0=\frac{x+4}{y}

    Can you do it from here??
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    You're expected to be able to recognise triangles that are of dimensions (1, 1, \sqrt{2}) and (1, \sqrt{3}, 2).

    Looking at the large triangle, its dimensions are

    (x, \sqrt{3}\,x, 2x).


    So the base of this shape is \sqrt{3}\,x.


    Using that information, you should be able to see that

    \sqrt{3}\,x - x = 4

    x(\sqrt{3} - 1) = 4

    x = \frac{4}{\sqrt{3} - 1}

    x = \frac{4(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}

    x = \frac{4(\sqrt{3} + 1)}{3 - 1}

    x = \frac{4(\sqrt{3} + 1)}{2}

    x = 2(\sqrt{3} + 1)

    x = 2\sqrt{3} + 2.
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