SOLUTION: Identify the theorem or postulate that is related to the measures of the angles in the pair, and find the unknown angle measures. m∠6=(30x+30)∘, m∠7=(45x−30)∘ The f

Algebra ->  Angles -> SOLUTION: Identify the theorem or postulate that is related to the measures of the angles in the pair, and find the unknown angle measures. m∠6=(30x+30)∘, m∠7=(45x−30)∘ The f      Log On


   



Question 1167444: Identify the theorem or postulate that is related to the measures of the angles in the pair, and find the unknown angle measures.
m∠6=(30x+30)∘, m∠7=(45x−30)∘
The figure shows two parallel lines and a transversal. The intersection of the first parallel line and a transversal forms four angles. The upper left angle is labeled as five, the upper right angle is labeled as one, the lower right angle is labeled as six, and the lower left angle is labeled as two. The intersection of the second parallel line and a transversal forms four angles. The upper left angle is labeled as seven, the upper right angle is labeled as three, the lower right angle is labeled as eight, and the lower left angle is labeled as four.

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
i believe you are talking about alternate interior angles formed by two lines intersected by a transversal.

if the two lines are parallel, then the alternate interior angles are congruent, i.e. equal.

angle 6 and 7 are alternate interior angles of parallel lines.

therefore, they are equal.

you get:

30x + 30 = 45x - 30
subtract 30x from both sides of the equation and add 30 to both sides of the equation to get:
60 = 15x
solve for x to get:
x = 60/15 = 4

angle 6 is equal to 30x + 30 = 120 + 30 = 150
angle 7 is equal to 45x - 30 = 180 - 30 = 150

both angles are equal, so the value of each angle is good.

angle 6 and angle 7 are each equal to 150 degrees.

here's a reference.

https://www.storyofmathematics.com/alternate-interior-angles#:~:text=Alternate%20interior%20angles%20are%20angles,opposite%20sides%20of%20the%20transversal.