Chapter 3, Problem 11CT

To determine

To show: the measures of the numbered angles with the given figure.

Answer to Problem 11CT

The measure of the numbered angles: $\angle 1={72}^{\circ },\angle 2={72}^{\circ },\angle 3={36}^{\circ },\angle 4={108}^{\circ },\angle 5={36}^{\circ }$ .

Explanation of Solution

Given information:

  $ABCDE$ is a regular polygon.

  

Calculation:

Considering the following figure:

In above figure line segment $ABCDE$ is a regular polygon.

The measure of $\angle CBE={72}^{\circ }$ (1)

Consider the following theorem

The theorem states that two non-adjacent interior angles on the opposite sides of transversal are consider as an alternate interior angle which are by definition equal to each other

So according to above figure if transversal $BC$ bisects two line segments $BE\parallel DF$ .Then $\angle CBE=\angle 2={72}^{\circ }$ because they are alternate interior angles

Hence, $\angle 2={72}^{\circ }$ (2)

Consider the following theorem

The theorem states that the two interior angles on the same side of the transversal are equal to each other

So according to above figure if transversal $BC$ bisects two line segments $BE\parallel DF$ Then $\angle 1=\angle 2={72}^{\circ }$ because they are same-side interior angles

Hence, $\angle 1={72}^{\circ }$ (3)

If in $\text{Δ}BCF$ measure of $\angle 1={72}^{\circ }$ and $\angle 2={72}^{\circ }$

Then measure of $\angle 3$ can be found by using the following theorem

The theorem states that sum of the measure of the angles of a triangle is ${180}^{\circ }$

Hence, solve as follows,

  $\angle 3={180}^{\circ }-(\angle 1+\angle 2)$

This means $\angle 3={180}^{\circ }-(\angle 1+\angle 2)$ where $\angle 3={180}^{\circ }-(\angle 1+\angle 2)$ and $\angle 1={72}^{\circ }$

Hence, solve as follows,

  $\angle 3={180}^{\circ }-(\angle 1+\angle 2)$

  $\angle 3={180}^{\circ }-\left({72}^{\circ }+{72}^{\circ }\right)$

  $\angle 3={36}^{\circ }$ (4)

Consider the following theorem

The theorem states that the two angles in corresponding position relative to the two lines are equal to each other

So according to above figure if transversal $BC$ bisects two line segments $BE\parallel DF$

Then $\angle 3=\angle 5={36}^{\circ }$ because they are two angles in corresponding position relative to the two lines $BE\parallel DF$

Hence, $\angle 5={36}^{\circ }$ (5)

Draw a perpendicular angle bisector $AG$ meet at $BE$ which bisects $\angle 4$ into two equal angles

  $\angle BAG=\angle EAG$

If in $\text{Δ}AGB$ measure of $\angle AGB={90}^{\circ }$ and $\angle 5={36}^{\circ }$

Then measure of $\angle BAG$ can be found by using the following theorem

The theorem $3-11$ sates that sum of the measure of the angles of a triangle is ${180}^{\circ }$

Hence, solve as follows,

  $\angle BAG={180}^{\circ }-(\angle AGB+\angle 5)$

This means $\angle BAG={180}^{\circ }-(\angle AGB+\angle 5)$

  $\angle AGB={90}^{\circ }$

Hence, solve as follows,

  $\angle 5={36}^{\circ }$

  $\begin{array}{l}\angle BAG={180}^{\circ }-(\angle AGB+\angle 5)\\ \angle BAG={180}^{\circ }-\left({90}^{\circ }+{36}^{\circ }\right)\\ ={180}^{\circ }-{126}^{\circ }\\ ={54}^{\circ }\end{array}$ .

If $\angle BAG={54}^{\circ }$ then $\angle BAG=\angle EAG={54}^{\circ }$

Thus

  $\angle 4=\angle BAG+\angle EAG$

  $\angle 4={54}^{\circ }+{54}^{\circ }$

  $\angle 4={108}^{\circ }$

Therefore, measure of the numbered angles:

  $\angle 1={72}^{\circ },\angle 2={72}^{\circ },\angle 3={36}^{\circ },\angle 4={108}^{\circ },\angle 5={36}^{\circ }$