Given that we assume that all the bases of the triangles are parallel.
We can use AAA or Angle-Angle-Angle to prove that these triangles are similar.
Each parallel line creates the same angle when intersecting with the same side.
For example:
The bases of each triangle cross the left side of all the triangle.
Each angle made by the intersecting of the the parallel base and the side are the same.
Thus, each corresponding angle of all the triangles are congruent.
If these angles are congruent, then we have similar triangles.
Answer:
Problem B: x = 12; m<EFG = 48
Problem C: m<G = 60; m<J = 120
Step-by-step explanation:
Problem B.
Angles EFG and IFH are vertical angles, so they are congruent.
m<EFG = m<IFH
4x = 48
x = 12
m<EFG = m<IFH = 48
Problem C.
One angle is marked a right angle, so its measure is 90 deg.
The next angle counterclockwise is marked 30 deg.
Add these two measures together, and you get 120 deg.
<J is vertical with the angle whose measure is 120 deg, so m<J = 120 deg.
Angles G and J from a linear pair, so they are supplementary, and the sum of their measures is 180 deg.
m<G = 180 - 120 = 60
Mmm, x is equals to -7....
Answer:
Step-by-step explanation:
Answer:18
Step-by-step explanation: 18=2.3.3
