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Math Help - Special right triangles

  1. #1
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    Special right triangles

    Hello Guys,

    Right here we have a hexagon. Each side is going to be 6 since 360/6 = 6

    Now.. I don't know how BF = 6SQRT3
    I tried drawing a altitude from Angle A to BF but that didn't work. I got 6SQRT for half of BF.

    and CF = 12 but I don't know how. I know it's a 30 60 90 special triangle.
    Attached Thumbnails Attached Thumbnails Special right triangles-screen-shot-2014-02-08-8.04.52-pm.png  
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  2. #2
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    Re: Special right triangles

    Hey Cake.

    I'll get you started by noting that ABF is a triangle with interior angle sum of 180 degrees and is an isosceles triangle with two of the sides being 6 cm.

    Since the interior angle sum is 720 degrees we know that angle FAB is equal to 720/6 = 120 degrees. We can use the cosine rule to get the length of BF by noting that:

    c^2 = a^2 + b^2 - 2abcos(C) (cosine rule). Note that 120 degrees = 2*pi/3 in radians.
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  3. #3
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    Re: Special right triangles

    Hey Chiro,
    Yes, I've noticed that. We can't use the cosine rule yet though. We just have to go based on what we know on special right triangles. Excluding cosine, sin, and tangent.
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  4. #4
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    Re: Special right triangles

    Quote Originally Posted by Cake View Post
    Hey Chiro,
    Yes, I've noticed that. We can't use the cosine rule yet though. We just have to go based on what we know on special right triangles. Excluding cosine, sin, and tangent.
    Not sure what you mean by special right triangle. Is a 30/60/90 right triangle special?

    You'll note that the interior angles of a hexagon are 120 degrees. Drop a perpendicular line from A to BF at Y. That line bisects A and BF.
    So you have a 30/60/90 triangle AYB with AB=6
    thus you know AY=3 and YB=3sqrt(3).

    Due to symmetry YF=YB and so BF=2YB = 6sqrt(3)

    BFC is a right triangle with legs 6sqrt(3) and 6. CF is the hypotenuse and thus equals

    sqrt(36*3 + 36) = sqrt(144) = 12

    now that you have the idea see if you can crack 36.
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  5. #5
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    Re: Special right triangles

    Hello, Cake!

    Darn! . . . LaTeX is not working!


    In regular hexagon ABCDEF, AB = 6 inches.
    Find the exact length of: (a) BF, (b) CF.

    Code:
              B   6   C
              * * * * *
             *:*     * *
           6* : *   *6  *6
           *  :  *O*     *
        A * * + * * * * * * D
           *  :  * *     *
           6* : *6  *   *6
             *:*     * *
              * * * * *
              F   6   E

    The hexagon is comprised of six equilateral triangles of side 6.
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ._
    The altitude of an equilateral triangle of side x is: (√3/2)x.
    . . . . . . . . . . . . . . . . . . . . . . . . _ . . . . . . . . _
    Hence, the altitude of ∆ABO is: (√3/2)(6) = 3√3.
    . . . . . . . . . . . . . . . ._
    Therefore: .BF .= .6√3 inches.


    And we can see that: .CF .= .6 + 6 .= .12 inches.
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