Chapter 3.2, Problem 16WE

To determine

To find: the values of x , y and z .

Answer to Problem 16WE

  $x=30$

  $y=8$

  $z=11$

Explanation of Solution

Given:

The diagram:

Concept Used:

Postulate 10: If two parallel lines are cut by a transversal then the corresponding angles are congruent

Theorem 3-2: If two parallel lines are cut by a transversal then the alternate interior angles are congruent.

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

Label the given figure as shown below:

In the above figure, lines $\overline{AB}\parallel \overline{CD}$ and $\overline{GH}$ is a transversal, so by theorem 3-3,

  $m\angle BQR+m\angle QRD=180$ (Same-side interior angles)

This implies that,

  $\begin{array}{c}3x+90=180\\ 3x=180-90\\ 3x=90\\ x=30\end{array}$

Now lines $\overline{AB}\parallel \overline{CD}$ and $\overline{EF}$ is a transversal, so by postulate 10,

  $\angle FPA\cong \angle PRC$ (Corresponding angles)

That is,

  $\begin{array}{c}m\angle PRC=m\angle FPA\\ 8y+4=68\\ 8y=68\end{array}$

  $\begin{array}{c}8y=64\\ y=\frac{64}{8}\\ y=8\end{array}$

Now, lines $\overline{AB}\parallel \overline{CD}$ and $\overline{GH}$ is a transversal, so by theorem 3-2,

  $\angle BQR\cong \angle QRC$ (Alternate interior angles)

That gives,

  $m\angle PRD=m\angle EAP$

But,

  $\angle QRC=\angle QRP+\angle PRC$

So,

  $m\angle QRC=m\angle QRP+m\angle PRC$

That is,

  $\begin{array}{c}m\angle QRC=m\angle BQR\\ m\angle QRP+m\angle PRC=m\angle BQR\\ 68+2z=90\\ 2z=90-68\end{array}$

  $\begin{array}{c}2z=22\\ z=\frac{22}{2}\\ z=11\end{array}$