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Math Help - Application of similar triangles

  1. #1
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    Application of similar triangles

    So I have this one problem:

    A wall is being supported by a beam passing over a dumpster 5 feet high and 4 feet from the wall. Find the length of the shortest possible beam.

    I know that you need to use similar triangles, but besides that I don't really know what to do. Please Help!!!!!!!!!
    This is what I have figured out:

    from the ground to the beam (along the wall) is v, and from the dumpster to the beam is v-5

    from the wall to the beam (along the ground) is u, and from the dumpster to the beam is u-4.
    the beam is x.

    I've got it all into one equation: x^2 = v^2 + (20 +4)^2
    (that's 20 over v-5) (v-5)

    I'm stuck now. PLEASE PLEASE PLEASE HELP!!!!!!!!!!!!!!!!!
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  2. #2
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    You can also do it with trig.

    Let {\theta} be the angle between the beam and the ground.

    The length of the beam is L=L_{1}+L_{2}

    L_{1}=length from ground to dumpster.

    L_{2}= length from dumpster to wall.

    L=5csc({\theta})+4sec({\theta})

    \frac{dL}{d{\theta}}=-5csc({\theta})cot({\theta})+4sec({\theta})tan({\th  eta}), \;\ 0<{\theta}<\frac{\pi}{2}

    =\frac{-5cos^{3}({\theta})+4sin^{3}({\theta})}{sin^{2}({\t  heta})cos^{2}({\theta})}

    Now, \frac{dL}{d{\theta}}=0 if 4sin^{3}({\theta})=5cos^{3}({\theta})

    Can you finish?.
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  3. #3
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    Hello, jahichuanna!

    This is a classic . . . The "shortest ladder" problem.


    A wall is being supported by a beam passing over a dumpster 5 ft high and 4 ft from the wall.
    Find the length of the shortest possible beam.
    It is easier using Trigonometry . . .
    Code:
                            * C
                          * |
                        *   |
                   B  * θ   |
                    * - - - * D
                  * |   4   |
                *   |       |
              *     |5      |
            * θ     |       |
        A * - - - - * - - - * E
                    F   4

    In right triangle BFA\!:\;\;\sin\theta \:=\:\frac{5}{AB} \quad\Rightarrow\quad AB \:=\:5\csc\theta

    In right triangle CDB\!:\;\;\cos\theta \:=\:\frac{4}{BC} \quad\Rightarrow\quad BC \:=\:4\sec\theta

    So we have: . AC \:=\:AB + BC \quad\Rightarrow\quad L \;=\;5\csc\theta + 4\sec\theta


    And that is the function we must minimize . . .

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  4. #4
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    If you must use similar triangles, then:

    y=height of wall and x=horizontal distance from ground to dumpster.

    \frac{5}{x}=\frac{y}{x+4}

    y=\frac{5(x+4)}{x}

    Now, use Pythagoras and minimize.

    You should get the same answer in both methods
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