Question

find the measure of each angle. (see example 3.)
5.

1. $angle uvw$ and $angle xyz$ are complementary angles, $mangle uvw=(x - 10)^{circ}$, and $mangle xyz=(4x-10)^{circ}$
2. $angle efg$ and $angle lmn$ are supplementary angles, $mangle efg=(3x + 17)^{circ}$, and $mangle lmn=(\frac{1}{2}x - 5)^{circ}$

use the diagram. (see example 4.)

1. identify all the linear pairs that include $angle1$.
2. identify all the linear pairs that include $angle7$.
3. are $angle6$ and $angle8$ vertical angles?
4. are $angle2$ and $angle5$ vertical angles?

Explanation:

Step1: Solve for x in problem 5

Since the sum of the angles on a straight - line is 180°, we set up the equation $(3x + 5)+(10x-7)=180$.
Combining like terms gives $13x - 2=180$.
Adding 2 to both sides: $13x=182$.
Dividing both sides by 13: $x = 14$.

Step2: Find the measure of each angle in problem 5

For the first angle: $3x + 5=3\times14 + 5=42 + 5=47^{\circ}$.
For the second angle: $10x-7=10\times14-7=140 - 7=133^{\circ}$.

Step3: Solve for x in problem 6

Since $\angle BAC$ is a right - angle ($90^{\circ}$), we set up the equation $(15x - 2)+(7x + 4)=90$.
Combining like terms gives $22x+2 = 90$.
Subtracting 2 from both sides: $22x=88$.
Dividing both sides by 22: $x = 4$.

Step4: Find the measure of each angle in problem 6

For the first angle: $15x - 2=15\times4-2=60 - 2=58^{\circ}$.
For the second angle: $7x + 4=7\times4 + 4=28 + 4=32^{\circ}$.

Step5: Solve for x in problem 7

Since $\angle UVW$ and $\angle XYZ$ are complementary ($m\angle UVW+m\angle XYZ = 90^{\circ}$), we set up the equation $(x - 10)+(4x-10)=90$.
Combining like terms gives $5x-20 = 90$.
Adding 20 to both sides: $5x=110$.
Dividing both sides by 5: $x = 22$.

Step6: Find the measure of each angle in problem 7

$m\angle UVW=x - 10=22-10 = 12^{\circ}$.
$m\angle XYZ=4x-10=4\times22-10=88 - 10=78^{\circ}$.

Step7: Solve for x in problem 8

Since $\angle EFG$ and $\angle LMN$ are supplementary ($m\angle EFG+m\angle LMN = 180^{\circ}$), we set up the equation $(3x + 17)+(\frac{1}{2}x-5)=180$.
Combining like terms: $3x+\frac{1}{2}x+17 - 5=180$, or $\frac{6x + x}{2}+12=180$, or $\frac{7x}{2}=168$.
Multiplying both sides by $\frac{2}{7}$: $x = 48$.

Step8: Find the measure of each angle in problem 8

$m\angle EFG=3x + 17=3\times48+17=144 + 17=161^{\circ}$.
$m\angle LMN=\frac{1}{2}x-5=\frac{1}{2}\times48-5=24 - 5=19^{\circ}$.

Step9: Answer problem 9

The linear pairs that include $\angle1$ are $\angle1$ and $\angle2$, $\angle1$ and $\angle5$.

Step10: Answer problem 10

The linear pairs that include $\angle7$ are $\angle7$ and $\angle6$, $\angle7$ and $\angle8$.

Step11: Answer problem 11

$\angle6$ and $\angle8$ are vertical angles.

Step12: Answer problem 12

$\angle2$ and $\angle5$ are not vertical angles.

Answer:

Problem 5: First angle: $47^{\circ}$, Second angle: $133^{\circ}$
Problem 6: First angle: $58^{\circ}$, Second angle: $32^{\circ}$
Problem 7: $m\angle UVW = 12^{\circ}$, $m\angle XYZ=78^{\circ}$
Problem 8: $m\angle EFG = 161^{\circ}$, $m\angle LMN=19^{\circ}$
Problem 9: $\angle1$ and $\angle2$, $\angle1$ and $\angle5$
Problem 10: $\angle7$ and $\angle6$, $\angle7$ and $\angle8$
Problem 11: Yes
Problem 12: No