Answer:
m<A = 54 degrees
m<AOB = 72 degrees
Step-by-step explanation:
First, if arcDA is 108 degrees, than m<DOA is 108 degrees
m<DOA and m<AOB are supplementary, so m<AOB=180-108=72 degrees
Next m<A
If this double triangle kind of helix is in the center of an ellipse, it means it is congruent. If it this happens in a circle it meant the triangles are both congruent and isosceles.
Hence, triAOB is isosceles.
SO, the angle of A and B are congruent and the equation is half of the supplement of AOB, in which case is DOA---
m<A=m<B=m<DOA/2
m<A=108/2
m<A=54 degrees
I also went to RSM,
so good luck for the rest of Eighth Grade Geometry.
