Posted by **Marisol** on Sunday, October 4, 2009 at 4:28pm.

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!

- Math -
**Reiny**, Sunday, October 4, 2009 at 4:40pm
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

- Math -
**Marisol**, Sunday, October 4, 2009 at 4:57pm
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?

- Math -
**Reiny**, Sunday, October 4, 2009 at 5:15pm
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

- Math -
**Marisol**, Sunday, October 17, 2010 at 12:53am
word

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