Asked 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!
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!
Answers
Answered by
Reiny
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
and x = 34
then 3x + angle BOC = 162
angle BOC = 162-3(34) = 102
Then angle DOA = 34+102+34 = 170
Answered by
Marisol
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?
Answered by
Reiny
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
Answered by
Marisol
word
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