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Math Help - Similar triangle problem

  1. #1
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    Nov 2009
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    Similar triangle problem

    ok...i tried searching a bit before posting this question, and it may be a little long winded so try to hang with me. this is going to be for an audio frequency tuning calculator i'm making in visual basic, so i need to come up with a formula to figure all 3 sides of similar triangles.

    ok...so, a user puts in 2 different depths, height, and width (tho width is irrelevant in this scenario). now, i know that subtracting the depths will give me one measurement (side a), and height entered is another measurement (side b). pythagorean theorem gives me the hypotenuse of the big triangle (side c). that there is the easy part. so for instance we'll say the bottom is 12" long, and height is 14" which makes the hypotenuse 18.44". so then we have this:



    now...the similar triangle lies within. the hypotenuse and side A are made of wood, .75" thick. that effectively gives us this:



    i can determine the height of the new triangle by subtracting .75" from the height, giving me 13.25". and put 14/13.25 = 12/z, i can figure that side z = 11.36". and pythagorean theorem to get the hypotenuse. getting me this:



    aaaaaaand this is where i get stuck. i need the sides of that smaller white triangle. distance from the big hypotenuse to the small one is .75" thick, but where it meets the other 2 edges is longer and i dont know how to find the sides of the new triangle


    so basically i need the lengths of m, n, and o. or any combination of the 2 and i can figure out the third. then figure a way to incorporate them into a formula for my program

    i hope i posted that as clear as possible. any help would be GREATLY appreciated. i've been stumped on this for a while now lol. thanks in advance!
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  2. #2
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    Lexington, MA (USA)
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    Hello, ExpoSport!

    The problem is a bit more complicated than that.

    I'll pick the dimensions for my convenience.

    Suppose the right triangle has sides: 12, 16, 20
    Code:
        A o
          |  *
          |     *
          |        *
          |           *
          |              *  20
       12 |                 *
          |                    *
          |                       *
          |                          *
          |                             *
          |                                *
        C o - - - - - - - - - - - - - - - - - o B
                           16
    Suppose we cut a 1-inch strip from the hypotenuse.

    The new (smaller) triangle is not 11 by 15 . . .


    Code:
        A o   E
          |  o
          |θ/   *
          |/ 1     *
        D o           *
          |  *           *
          |     *           *
          |        *           *
          |           *           *   G
          |              *           o
          |                 *      1/   *
          |                    *   /     θ *
        C o - - - - - - - - - - - o - - - - - o B
                                  F

    Let \theta = \angle B,\:\text{ then: }\angle ADE = \theta

    Then: . \begin{array}{c}\sin\theta \:=\:\dfrac{12}{20} \:=\:\dfrac{3}{5} \\ \\[-3mm] \cos\theta \:=\:\dfrac{16}{20} \:=\:\dfrac{4}{5} \end{array}

    DE = FG = 1 is the width of the strip.


    In right triangle FGB\!:\;\;\sin\theta \:=\:\frac{1}{FB} \quad\Rightarrow\quad \frac{3}{5} \:=\:\frac{1}{FB} \quad\Rightarrow\quad FB \:=\:\frac{5}{3}

    In right triangle AED\!:\;\;\cos\theta \:=\:\frac{1}{AD} \quad\Rightarrow\quad \frac{4}{5} \:=\:\frac{1}{AD} \quad\Rightarrow\quad AD \:=\:\frac{5}{4}


    Hence: . \begin{array}{ccccccc}CF \:=\:16 - \frac{5}{3} \:=\:14\frac{1}{3} \\ \\[-3mm] CD \:=\:12 - \frac{5}{4} \:=\:10\frac{3}{4} \end{array}


    Get the idea?

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  3. #3
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    Nov 2009
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    nice!! that's exactly what i'm looking for. and haha yes...that's a lil more complicated than i thought. i thought i may be getting close, but had a feeling there was a lot more to it to figure out what i needed. so...following that method, and using the dimensions for my triangle, i have:

    <br /> <br />
\begin{array}{c}\sin\theta \:=\:\dfrac{14}{18.44} \:=\:\dfrac{7}{9.22} \\ \\[-3mm] \cos\theta \:=\:\dfrac{12}{18.44} \:=\:\dfrac{6}{9.22} \end{array}<br />

    and i use .75 instead of 1, so:

    <br /> <br />
FGB\!:\;\;\sin\theta \:=\:\frac{.75}{FB} \quad\Rightarrow\quad \frac{7}{9.22} \:=\:\frac{.75}{FB} \quad\Rightarrow\quad FB \:=\:\frac{6.915}{7}<br />

    <br /> <br />
AED\!:\;\;\cos\theta \:=\:\frac{.75}{AD} \quad\Rightarrow\quad \frac{6}{9.22} \:=\:\frac{.75}{AD} \quad\Rightarrow\quad AD \:=\:\frac{6.915}{6}<br />

    to get:

    <br /> <br />
\begin{array}{ccccccc}CF \:=\:12 - \frac{6.915}{7} \:=\:11.01 \\ \\[-3mm] CD \:=\:14 - \frac{6.915}{6} \:=\:12.84 \end{array}<br />

    does that look correct?
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