91Ó°ÊÓ

The figure can be drawn as below:

Let$∠AOB=x$.The angle bisector divides the angle into two equal parts.

As$\stackrel{¯}{OC}$bisects$∠AOB$. Therefore,$∠AOC=∠BOC=\frac{x}{2}$.

As$\stackrel{¯}{OD}$bisects $∠AOC=\frac{x}{2}$. Therefore, $∠AOD=∠DOC=\frac{x}{4}$.

As$\stackrel{¯}{OE}$bisects $∠AOD=\frac{x}{4}$. Therefore, $∠AOE=∠EOD=\frac{x}{8}$.

As$\stackrel{¯}{OF}$bisects $∠AOE=\frac{x}{8}$. Therefore, $∠AOF=∠FOE=\frac{x}{16}$.

In order to find x, observe that$∠BOF=∠AOB−∠AOF$ and$∠BOF=120$.

$$\begin{gathered} ∠AOB−∠AOF=120 \\ x−\frac{x}{16}=120 \\ \frac{\left(16x−x\right)}{16}=120 \\ \frac{15x}{16}=120 \\ \frac{15x}{16}=120 \\ x=\frac{1920}{15} \\ =128 \end{gathered}$$

Finally, to find$∠DOE$, substitute 128 for x into $∠DOE=\frac{x}{8}$.

$$\begin{gathered} ∠DOE=\frac{x}{8} \\ =\frac{128}{8} \\ =16 \end{gathered}$$

Therefore, the value of the angle is $∠DOE=16$.

The figure can be drawn as below:

Let $∠AOC=x$.The angle bisector divides the angle into two equal parts.

As $\stackrel{¯}{OD}$bisects$∠AOC$

Therefore,

$$\begin{gathered} ∠AOD=∠DOC \\ =\frac{x}{2} \end{gathered}$$

As $\stackrel{¯}{OE}$bisects$∠AOD$

$$\begin{gathered} ∠AOE=∠EOD \\ =\frac{\frac{x}{2}}{2} \\ ∠AOE=\frac{x}{4} \end{gathered}$$

As $\stackrel{¯}{OF}$bisects$∠AOE$

Therefore,

$$\begin{gathered} ∠AOF=∠FOE \\ =\frac{\frac{x}{4}}{2} \\ ∠AOF=\frac{x}{8} \end{gathered}$$

The equation becomes;

$$\begin{gathered} ∠FOC=∠AOC−∠AOF \\ ∠FOC=x−\frac{x}{8} \\ ∠FOC=\frac{7x}{8} \end{gathered}$$

As $\stackrel{¯}{OG}$bisects$∠FOC$

Therefore,

$$\begin{gathered} ∠COG=∠FOG \\ =\frac{\frac{7x}{8}}{2} \\ ∠COG=\frac{7x}{16} \end{gathered}$$

It is given that $∠COG=35$and $∠COG=\frac{7x}{16}$, therefore,

width="74">7x16=35x=5607x=80

Substitute 80 for x to find the value of $∠AOF$and $∠AOE$.

width="96" height="114" role="math">∠AOF=x8=808=10

And,

$$\begin{gathered} ∠AOE=\frac{x}{4} \\ =\frac{80}{4} \\ =20 \end{gathered}$$

The equation becomes;

width="186" height="68" role="math">∠AOG=∠AOF+∠FOG=10+35=45

And the other equation is;

$$\begin{gathered} ∠EOG=∠AOG−∠AOE \\ \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}=45−20 \\ \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}=25 \end{gathered}$$

Therefore, the value is $∠EOG=25$.