Question

name: 2 - 3 additional practice parallel lines and triangle angle sums for exercises 1 - 6, find the value of each variable. 1. 63°, 92°, x° 2. 20°, 87°, x° 3. 80°, 40°, x° 4. 85°, 28°, 15°, x°, y°, z° 5. 43°, 24°, 57°, x°, y°, z° 6. 84°, 50°, 11°, x°, y°, z°

Explanation:

Step1: Recall triangle - angle sum property

The sum of interior angles of a triangle is 180°.

Step2: Solve for Exercise 1

For the triangle with angles 63°, 92°, and \(x\):
\(x = 180-(63 + 92)=180 - 155 = 25\)

Step3: Solve for Exercise 2

For the first - triangle with angles 20°, 87°, and \(x\):
\(x=180-(20 + 87)=180 - 107 = 73\)

Step4: Solve for Exercise 3

For the triangle with angles 40°, 80°, and \(x\):
\(x = 180-(40 + 80)=180 - 120 = 60\)

Step5: Solve for Exercise 4

First, find the non - labeled angle in the small triangle with 28° and 15°. Let this angle be \(a\), then \(a=180-(28 + 15)=137\).
In the large triangle with 85° and \(a = 137\):
\(x=180-(85 + 137)=180 - 222\) (This is wrong. Let's use another approach. The exterior - angle property: The exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles).
The exterior angle of the small triangle is \(28+15 = 43\).
In the large triangle, \(x + 85+43=180\), so \(x=180-(85 + 43)=52\).
\(y\) and the angle with measure 28° are vertical angles, so \(y = 28\).
\(z=180-(52 + 28)=100\)

Step6: Solve for Exercise 5

The exterior angle of the small triangle is \(24 + 43=67\).
In the large triangle, \(x+57 + 67=180\), so \(x=180-(57 + 67)=56\).
\(y\) and 43° are vertical angles, so \(y = 43\).
\(z=180-(56 + 43)=81\)

Step7: Solve for Exercise 6

The exterior angle of the small triangle is \(50+11 = 61\).
In the large triangle, \(x+84+61=180\), so \(x=180-(84 + 61)=35\).
\(y\) and 50° are vertical angles, so \(y = 50\).
\(z=180-(35 + 50)=95\)

Answer:

1. \(x = 25\)
2. \(x = 73\)
3. \(x = 60\)
4. \(x = 52\), \(y = 28\), \(z = 100\)
5. \(x = 56\), \(y = 43\), \(z = 81\)
6. \(x = 35\), \(y = 50\), \(z = 95\)