Math
posted by Marisol .
B is in the interior of <AOC. C is in the interior of <BOD. D is in the interior of <COE. m<AOE=162, m<COE=68, and m<AOB=m<COD=m<DOE. Find m<DOA.
I have a sketch of the angles, and this is what I have so far as for work:
162=68+3x
94=3x
x=31.3
x is the measure of the three equal angles, but not DOA. I'm not sure what to do after this. Help?
Thanks!

According to your description in my diagram AD bisects angle COE, so 2x = 68
and x = 34
then 3x + angle BOC = 162
angle BOC = 1623(34) = 102
Then angle DOA = 34+102+34 = 170 
Wait, I'm confused. How does the angle equal more than what the entire thing is, which is m<AOE=162? Or did I sketch this wrong?

You are right for catching that!
My error is in the second last line
angle BOC = 1623(34) = 60 , (not 102)
and
Then angle DOA = 34+60+34 = 128 
word
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