Chapter 1.4, Problem 36WE

(a)

To determine

To evaluate: The value of $\angle DOE$

Answer to Problem 36WE

16°

Explanation of Solution

Given information:

The figure can be drawn as shown below:

  

Formula used:

The bisector of an angle property.

Calculation:

To find the angle by using definition of bisector of an angle:

Take $\angle AOB=x$

As $\overline{OC}$ bisects $\angle AOB$

Therefore,

  $\begin{array}{l}\angle AOC=\angle BOC\\ =\frac{x}{2}\end{array}$

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

The angle bisector divides the angle into two equal parts.

Therefore,

  $\begin{array}{l}\angle AOD=\angle DOC\\ =\frac{\frac{x}{2}}{2}\\ \angle AOD=\frac{x}{4}\end{array}$

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

Therefore,

  $\begin{array}{l}\angle AOE=\angle EOD\\ =\frac{\frac{x}{4}}{2}\\ \angle AOE=\frac{x}{8}\end{array}$

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

Therefore,

  $\begin{array}{l}\angle AOF=\angle FOE\\ =\frac{\frac{x}{8}}{2}\\ \angle AOF=\frac{x}{16}\end{array}$

It is known that;

  $\angle BOF=120$

Also,

  $\angle BOF=\angle AOB-\angle AOF$

Therefore,

  $\angle AOB-\angle AOF=120$

Substitute the values;

  $\begin{array}{l}x-\frac{x}{16}=120\\ \frac{\left(16x-x\right)}{16}=120\\ \frac{15x}{16}=120\\ \frac{15x}{16}=120\end{array}$

That is,

  $\begin{array}{l}x=\frac{1920}{15}\\ x=128\end{array}$

Therefore,

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

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

Conclusion:

The value of the angle is $\angle DOE=16$

(b)

To determine

To calculate: The value of $\angle EOG$

Answer to Problem 36WE

25°

Explanation of Solution

Given information:

The figure can be drawn as shown below:

  

Formula used:

The bisector of an angle property.

Calculation:

Consider that $\angle AOC=x$

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

Therefore,

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

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

Therefore,

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

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

Therefore,

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

The equation becomes;

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

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

Therefore,

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

It is known that;

  $\angle COG=35$

Also, $\angle COG=\frac{7x}{16}$

Therefore,

  $\begin{array}{l}\frac{7x}{16}=35\\ x=\frac{560}{7}\\ x=80\end{array}$

Therefore,

  $\begin{array}{l}\angle AOF=\frac{x}{8}\\ \angle AOF=\frac{80}{8}\\ \angle AOF=10\end{array}$

And

  $\begin{array}{l}\angle AOE=\frac{x}{4}\\ \angle AOE=\frac{80}{4}\end{array}$

That is

  $\angle AOE=20$

The equation becomes;

  $\begin{array}{l}\angle AOG=\angle AOF+\angle FOG\\ \angle AOG=10+35\\ \angle AOG=45\end{array}$

And the other equation is;

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

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

Conclusion:

The value is $\angle EOG=25$