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Math Help - Geometry Problem...

  1. #1
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    Geometry Problem...

    Unfortunately, I need help with this problem by the end of tonight. If you think you have it, please post and show alllllll work.

    If ray OC bisects angle AOB, ray OD bisects angle AOC, ray OE bisects angle AOD, ray OF bisects angle AOE and ray OG bisects angle FOC.
    a. If m<BOF = 120, then, M<DOE = ___?
    b. If m<COG = 35, then m<EOG = ___?
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let m(\widehat{AOB})=\alpha.
    a) Then \displaystyle m(\widehat{BOC})=m(\widehat{COA})=\frac{\alpha}{2}
    \displaystyle m(\widehat{COD})=m(\widehat{DOA})=\frac{\alpha}{4}
    \displaystyle m(\widehat{DOE})=m(\widehat{EOA})=\frac{\alpha}{8}
    \displaystyle m(\widehat{EOF})=m(\widehat{FOA})=\frac{\alpha}{16  }
    We have \displaystyle m(\widehat{BOA})=m(\widehat{BOF})+m(\widehat{FOA})  \Rightarrow \alpha=120^{\circ}+\frac{\alpha}{16}\Rightarrow \alpha =128^{\circ}
    \displaystyle m(\widehat{DOE})=\frac{\alpha}{8}=16^{\circ}
    b) m(\widehat{BOA})=m(\widehat{BOC})+m(\widehat{COF})  +m(\widehat{FOA})\Rightarrow \alpha =\frac{\alpha}{2}+70^{\circ}+\frac{\alpha}{16}\Rig  htarrow \alpha=160^{\circ}
    \displaystyle m(\widehat{AOG}=35^{\circ}+\frac{\alpha}{16}=45^{\  circ}
    \displaystyle m(\widehat{EOG})=m(\widehat{AOG})-m(\widehat{AOE})=45^{\circ}-20^{\circ}=25^{\circ}
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  3. #3
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    Hello, icebreaker09!

    Did you make a sketch?


    If ray OC bisects angle AOB, ray OD bisects angle AOC, ray OE bisects angle AOD,
    ray OF bisects angle AOE and ray OG bisects angle FOC.

    a) If \angle BOF = 120^o, find \angle DOE

    b) If \angle COG = 35^o, find \angle EOG
    Let \angle AOB = \theta
    The diagram (partially drawn) looks like this . . .
    Code:
    B                               C
       *            θ/2            *                    * D
          *                       *      θ/4        *
             *                   *              *   θ/8
                *               *           *             * E
                   *           *        *         *
                      *       *     *       *
                         *   *  *   *
                            *   *   *   *   *   *   *   *   * A
                            O

    a) We are told that \angle BOF = 120^o
    Then: . \angle BOF \;=\;\angle BOC + \angle COD + \angle DOE + \angle EOF \;=\;\frac{\theta}{2} + \frac{\theta}{4} + \frac{\theta}{8} + \frac{\theta}{16} \;=\;\frac{15}{16}\theta
    Hence: . \frac{15}{16}\theta\;=\;120^o\quad\Rightarrow\quad  \theta \,=\,128^o
    Therefore: . \angle DOE \;=\;\frac{\theta}{8}\;=\;\frac{128^o}{8} \;=\;\boxed{16^o}


    b) We are told that: . \angle COG = 35^o
    Then: . \angle COG \;=\;\angle COD + \angle DOE + \angle EOF + \angle FOG \;=\;\frac{\theta}{4} + \frac{\theta}{8} + \frac{\theta}{16} + \frac{\theta}{32}\;=\;\frac{15}{32}\theta
    Hence: . \frac{15}{32}\theta\:=\:35^o\quad\Rightarrow\quad\  theta \,=\,\frac{224}{3}
    Since \angle EOG \;=\;\angle EOF + \angle FOG \;=\;\frac{\theta}{16} + \frac{\theta}{32} \;=\;\frac{3}{32}\theta
    . . we have: . \angle EOG \;=\;\frac{3}{32}\left(\frac{224}{3}\right) \;=\;\boxed{7^o}

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