Post a New Question

Math

posted by on .

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 - ,

    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 - ,

    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 - ,

    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 - ,

    word

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question