91Ó°ÊÓ

Arrowheads denotes line segment $\stackrel{¯}{AB}$is parallel to$\stackrel{¯}{CD}$

Line segment $\stackrel{¯}{EF}$is transversal to the parallel lines

Line segment $\stackrel{¯}{GH}$is perpendicular to the parallel lines

$∠BQR$ and $∠QRD$forms a pair of same-side interior angles

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

$m∠BQR+m∠QRD=180$

$3x+90=180$

$3x=180−90$

$x=\frac{90}{3}$

$x=30$

Thus, $x=30$

$∠APE$and $∠CRP$forms a pair of corresponding angles

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

$m∠CRP=m∠APE$

$8y+4=68$

$8y=68−4$

$y=\frac{64}{8}$

$y=8$

Thus$y=8$

$\stackrel{¯}{GH}$is perpendicular to $\stackrel{¯}{CD}$

$m∠CRQ=90$

$8y+4+2z=90$

$2z=90−4−8\left(8\right)$

$2z=86−68$

$z=\frac{22}{2}$

$z=11$

Thus $z=11$