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 = 162-3(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 = 162-3(34) = 60 , (not 102)
and
Then angle DOA = 34+60+34 = 128
word
To find the measure of angle DOA, we can start by observing that angle AOC is the sum of angles AOE and COE. So we have:
m<AOC = m<AOE + m<COE
m<AOC = 162 + 68
m<AOC = 230
Since angle AOB and angle COD are congruent, they have the same measure. Let's call that measure x:
m<AOB = m<COD = x
Given that the measures of these angles are equal, and the fact that three of them (x + x + x) add up to 230, we can set up the following equation:
3x = 230
Now, solve for x:
x = 230/3
x ≈ 76.67
Since x represents the measure of angle AOB and angle DOA, we can find the measure of angle DOA by subtracting the angles AOE and AOB from the angle AOC:
m<DOA = m<AOC - m<AOB
m<DOA = 230 - 76.67
m<DOA ≈ 153.33
Thus, the measure of angle DOA is approximately 153.33 degrees.