Question

geometry unit 3 lesson 7 homework
name khloe augustin date 9 - 24 - 25 period 5th
use the following information for questions 1 - 3. fsu way, a transversal, tj, crosses the streets cane avenue, unf lane, and gator alley.
determine the value of the variables and each angle measure given cane avenue, unf lane, and gator alley are parallel to each other.

1. ∠tln=(3x - 18)°, ∠mzq=(5x + 14)°, and ∠nlm = y°
2. ∠jzq=(9x + 9)° and ∠tln=(8x + 19)°
3. ∠zmr=(x + 14)°, ∠lmo=(2x - 6)°, and ∠lmr = y°

Explanation:

Step1: Analyze angles in question 1

Since $\angle TLN=(3x - 18)^{\circ}$ and $\angle MZQ=(5x + 14)^{\circ}$ are same - side interior angles and the lines are parallel, they are supplementary. So, $3x-18 + 5x+14=180$.
Combining like - terms gives $8x-4 = 180$.
Adding 4 to both sides: $8x=184$.
Dividing both sides by 8: $x = 23$.
$\angle MZQ=5x + 14=5\times23+14=115 + 14=129^{\circ}$.
$\angle TLN=3x-18=3\times23-18=69 - 18=51^{\circ}$.
Since $\angle NLM$ and $\angle MZQ$ are corresponding angles, $y = 129$.

Step2: Analyze angles in question 2

Since $\angle JZQ=(9x + 9)^{\circ}$ and $\angle TLN=(8x + 19)^{\circ}$ are corresponding angles and the lines are parallel, $9x+9=8x + 19$.
Subtracting $8x$ from both sides gives $x+9=19$.
Subtracting 9 from both sides: $x = 10$.
$\angle JZQ=9x + 9=9\times10+9=99^{\circ}$.
$\angle TLN=8x + 19=8\times10+19=99^{\circ}$.

Step3: Analyze angles in question 3

Since $\angle ZMR=(x + 14)^{\circ}$ and $\angle LMO=(2x - 6)^{\circ}$ are vertical angles, they are equal. So, $x + 14=2x-6$.
Subtracting $x$ from both sides gives $14=x - 6$.
Adding 6 to both sides: $x = 20$.
$\angle ZMR=x + 14=20+14=34^{\circ}$.
$\angle LMO=2x-6=2\times20-6=34^{\circ}$.
Since $\angle LMR$ is a straight - angle (180°), $y=180-(x + 14+2x - 6)=180-(3x + 8)$.
Substituting $x = 20$ gives $y=180-(3\times20 + 8)=180-(60 + 8)=112^{\circ}$.

Answer:

1. $x = 23$, $\angle TLN=51^{\circ}$, $\angle MZQ=129^{\circ}$, $y = 129$
2. $x = 10$, $\angle JZQ=99^{\circ}$, $\angle TLN=99^{\circ}$
3. $x = 20$, $\angle ZMR=34^{\circ}$, $\angle LMO=34^{\circ}$, $y = 112$