1. ## 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 = ___?

2. 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) $m(\widehat{BOA})=m(\widehat{BOC})+m(\widehat{COF}) +m(\widehat{FOA})\Rightarrow \alpha =\frac{\alpha}{2}+70^{\circ}+\frac{\alpha}{16}\Rig htarrow \alpha=160^{\circ}$
$\displaystyle m(\widehat{AOG}=35^{\circ}+\frac{\alpha}{16}=45^{\ circ}$
$\displaystyle m(\widehat{EOG})=m(\widehat{AOG})-m(\widehat{AOE})=45^{\circ}-20^{\circ}=25^{\circ}$

3. 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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# ray oc is the bisector of angle aob and od is ray

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