Q36. - Geometry

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 $m\angle EOG=\underset{¯}{?}$.

Step-by-Step Solution

Answer

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

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

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 ∠ D O E .

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" style="max-width: none; vertical-align: -56px;" $$\begin{gathered} \frac{7x}{16}=35 \\ x=\frac{560}{7} \\ x=80 \end{gathered}$$ 

 

Part b. Step 4. Calculate ∠ E O G

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

width="96" height="114" style="max-width: none; vertical-align: -58px;" $$\begin{gathered} \angle AOF=\frac{x}{8} \\ \begin{array}{c}=\frac{80}{8}\\ =10\end{array} \end{gathered}$$ 

 

 And,

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

 

 The equation becomes;

 width="186" height="68" style="max-width: none; vertical-align: -35px;" $$\begin{gathered} \angle AOG=\angle AOF+\angle FOG \\ =10+35 \\ =45 \end{gathered}$$

 

And the other equation is;

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

 

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