According to your description in my diagram AD bisects angle COE, so 2x = 68
and x = 34
then 3x + angle BOC = 162
angle BOC = 162-3(34) = 102
Then angle DOA = 34+102+34 = 170
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!
4 answers
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 = 162-3(34) = 60 , (not 102)
and
Then angle DOA = 34+60+34 = 128
My error is in the second last line
angle BOC = 162-3(34) = 60 , (not 102)
and
Then angle DOA = 34+60+34 = 128
word