Question

$\overrightarrow{\mathit{O}\mathit{C}}$bisects $\underrightarrow{\angle AOB}$, $\overrightarrow{OD}$bisects $\angle AOC$, $\overrightarrow{OE}$bisects $\angle AOD$, $\overrightarrow{OF}$bisects $\angle AOE$and$\overrightarrow{OG}$bisects $\angle FOC$.

1. If $m\angle BOF=120$, then $m\angle DOE=\underset{¯}{?}.$
2. If $m\angle COG=35$, then $m\angle EOG=\underset{¯}{?}$.

Short answer

a.If$m\angle BOF=120$, then $m\angle DOE=16$.

b.If$m\angle COG=35$, then $m\angle EOG=25$.

Step-by-step solution

Part a. Step 1. Draw the diagram.

The figure can be drawn as below:

Part a. Step 2. Apply bisector of an angle property.

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

As$\overline{OC}$bisects$\angle AOB$. Therefore,$\angle AOC=\angle BOC=\frac{x}{2}$.

As$\overline{OD}$bisects $\angle AOC=\frac{x}{2}$. Therefore, $\angle AOD=\angle DOC=\frac{x}{4}$.

As$\overline{OE}$bisects $\angle AOD=\frac{x}{4}$. Therefore, $\angle AOE=\angle EOD=\frac{x}{8}$.

As$\overline{OF}$bisects $\angle AOE=\frac{x}{8}$. Therefore, $\angle AOF=\angle FOE=\frac{x}{16}$.

Part a. Step 3. Calculate x.

In order to find x, observe that$\angle BOF=\angle AOB-\angle AOF$ and$\angle BOF=120$.

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

Part a. Step 4. Calculate ∠DOE.

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

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

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

Part b. Step 1. Draw the diagram.

The figure can be drawn as below:

Part b. Step 2. Apply bisector of an angle property.

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

As $\overline{OD}$bisects$\angle AOC$

Therefore,

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

As $\overline{OE}$bisects$\angle AOD$

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

As $\overline{OF}$bisects$\angle AOE$

Therefore,

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

The equation becomes;

$$\begin{gathered} \angle FOC=\angle AOC-\angle AOF \\ \angle FOC=x-\frac{x}{8} \\ \angle FOC=\frac{7x}{8} \end{gathered}$$

As $\overline{OG}$bisects$\angle FOC$

Therefore,

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

Part b. Step 3. Calculate x.

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

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

Part b. Step 4. Calculate ∠EOG

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

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

And,

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

The equation becomes;

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

And the other equation is;

$$\begin{gathered} \angle EOG=\angle AOG-\angle AOE \\ =45-20 \\ =25 \end{gathered}$$

Therefore, the value is $\angle EOG=25$.