Can you help me put this in least to greater
2 answers:
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Answer:
Part A
x is 46°
Part B
Alternate angles are angles that are in relatively opposite locations relative to a transversal
Please see attached diagram showing alternate angles
Step-by-step explanation:
Part A
∠DRP = 110° (Given)
∠QPA = 64° (Given)
∠QPR =
Given that AB is parallel to CD, we have;
∠DRP is congruent to ∠APR (Alternate angles to a transversal RP of parallel lines AB and CD)
Therefore, ∠APR = 115°
∠APR = ∠QPA + ∠QPR (Angle addition postulate)
∴ 115° = 64° + ∠QPR
∠QPR = 110° - 64° = 46°
x = 46°
Part B.
Given that AB is parallel to CD, the lines common (that intersects) both lines are the transversal lines
The angles formed between the parallel lines and the transversal lines have special relationships based on their position with respect to each other
In the question, the angle 110° given between CD and the transversal RP, is found to at an alternate position to the angle ∠APR between the same transversal RP and AB and given that alternate angles are always equal, angle ∠APR is therefore also equal to 110°.
