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Thread: Roof Structure Question find lengths

  1. #1
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    Roof Structure Question find lengths


    Find the rise of BD?
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  2. #2
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    I give you a hint.

    The area of Triangle ABD = area ABC +area BCD

    Now, call $\displaystyle BD=h$, the height.

    Then you can find,
    $\displaystyle CD=\sqrt{4700^2-h^2}$

    Thus,
    area ABD=$\displaystyle (1/2)h(3300 +\sqrt{4700^2-h^2})$

    area ABC can be computed by Heron's Formula

    area BCD=$\displaystyle (1/2)h\sqrt{4700^2-h^2}$

    You have an equation to solve for!
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  3. #3
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    Hello, Zak!

    In $\displaystyle \Delta ABC$, use the Law of Cosines to find $\displaystyle \angle ACB.$

    $\displaystyle \cos(\angle ACB)\:=\:\frac{3300^2 + 4700^2 - 7000^2}{2(3300)(4700)}\:=\:-0.516441006$

    Hence: .$\displaystyle \angle ACB \:\approx\:121.1^o$

    Then: .$\displaystyle \angle BCD\:=\:180^o - 121.1^o \:=\:58.9^o$


    In right triangle $\displaystyle BDC\!:\;\;\sin 58.9^o \:=\:\frac{BD}{4700}$

    Therefore: .$\displaystyle BD \:=\:4700\sin58.9^o \:=\:4024.717024 \:\approx\:4025$

    Last edited by Soroban; Dec 3rd 2006 at 05:51 PM.
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  4. #4
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    Hello, Zak!

    An alternate approach . . .

    In $\displaystyle \Delta ABC$, use the Law of Cosines to find angle $\displaystyle A.$

    $\displaystyle \cos A\:=\:\frac{7000^2 + 3300^2 - 4700^2}{2(7000)(3300)} \:=\:0.8181818181$

    . . Hence: .$\displaystyle A \:\approx\:35.1^o$

    In right triangle $\displaystyle BDA\!:\;\sin 35.1^o \:= \:\frac{BD}{7000}\quad\Rightarrow\quad BD \:=\:7000\sin35.1^o \:=\:4025.036764$

    . . Therefore: .$\displaystyle BD \:\approx\:4025$

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