Chapter 3.3, Problem 18WE

To determine

To find: the values of x and y that makes the red lines parallel and the blue lines parallel.

Answer to Problem 18WE

  $x=90$

  $y=130$

Explanation of Solution

Given:

The diagram:

Concept Used:

Theorem 3-6: If two lines are cut by a transversal and same-side interior angles are supplementary, then the lines are parallel.

Labeling the given figure:

From the above figure, it can be observed that the angles $\angle DAB$ and $\angle ABC$ are the same-side interior angles which are formed when the lines $\overline{AD}$ and $\overline{BC}$ are cut by the transversal $\overline{AB}$ .

Now by theorem 3-6, for the blue lines $\overline{AD}$ and $\overline{BC}$ to be parallel, we must have that the angles $\angle DAB$ and $\angle ABC$ are supplementary, that is,

  $\begin{array}{c}m\angle DAB+m\angle ABC=180\\ \left(x-40\right)+\left(x+40\right)=180\\ x-40+x+40=180\\ x+x-40+40=180\end{array}$

  $\begin{array}{c}2x=180\\ x=\frac{180}{2}\\ x=90\end{array}$

Now, if $x=90$ then $m\angle DAB$ is given by:

  $\begin{array}{c}m\angle DAB=x-40\\ =90-40\\ =50\end{array}$

Now, it can also be observed that the angles $\angle CDA$ and $\angle DAB$ are the same-side interior angles which are formed when the lines $\overline{AB}$ and $\overline{CD}$ are cut by the transversal $\overline{AD}$ .

Now by theorem 3-6, for the blue lines $\overline{AB}$ and $\overline{CD}$ to be parallel, we must have that the angles $\angle CDA$ and $\angle DAB$ are supplementary, that is,

  $\begin{array}{c}m\angle DAB+m\angle CDA=180\\ 50+y=180\\ y=180-50\\ y=130\end{array}$