Question

1. 8x + 12 in. 12x - 18 in. 12. 7x - 4° 5x + 8° 13. 13x + 2° 15x - 2° 14. 40° 3x + 20° 15. use what you know about angle measures to find x, y, or z. 28° 100°

Explanation:

Step1: Solve for question 11

Since the triangle is isosceles (two equal base - angles), we set the two non - base sides equal.
$8x + 12=12x-18$
Subtract $8x$ from both sides: $12 = 4x-18$.
Add 18 to both sides: $4x=30$, so $x = 7.5$.

Step2: Solve for question 12

In a right - triangle, the sum of the two non - right angles is $90^{\circ}$.
$(7x - 4)+(5x + 8)=90$
Combine like terms: $12x+4 = 90$.
Subtract 4 from both sides: $12x=86$, so $x=\frac{43}{6}\approx7.17$.

Step3: Solve for question 13

The two angles are vertical angles, so they are equal.
$13x + 2=15x-2$
Subtract $13x$ from both sides: $2 = 2x-2$.
Add 2 to both sides: $2x = 4$, so $x = 2$.

Step4: Solve for question 14

The two angles are corresponding angles (assuming parallel lines), so they are equal.
$3x + 20=40$
Subtract 20 from both sides: $3x=20$, so $x=\frac{20}{3}\approx6.67$.

Step5: Solve for question 15

First, $x$ and the $100^{\circ}$ angle are vertical angles, so $x = 100^{\circ}$.
The angle adjacent to the $28^{\circ}$ angle (vertically opposite) and $y$ and $100^{\circ}$ form a straight - line, so the angle adjacent to $28^{\circ}$ is $28^{\circ}$ (vertical angles).
$y+28 + 100=180$ (angles on a straight - line).
$y=180-(100 + 28)=52^{\circ}$.
$z$ and the angle adjacent to $28^{\circ}$ are corresponding angles (assuming parallel lines), so $z = 28^{\circ}$.

Answer:

1. $x = 7.5$
2. $x=\frac{43}{6}$
3. $x = 2$
4. $x=\frac{20}{3}$
5. $x = 100^{\circ}$, $y = 52^{\circ}$, $z = 28^{\circ}$