Posted by **Morgan** on Wednesday, August 27, 2008 at 8:32pm.

-> OB bisects <AOC

m<AOB=6x, m<AOC=2x+2y, m<BOC=y+5

- Honors Geometry -
**RCisME**, Friday, August 29, 2008 at 8:08pm
1. If OB bisects <AOC, then <AOB + <BOC = <AOC.

6x + y + 5 = 2x + 2y

solve for y and you get 4x + 5 = y

2. Also, since OB bisect <AOC, then m<AOB = m<BOC

6x = y + 5

Now, use what y equals from step one and subsitute it in so that:

6x = (4x + 5) + 5

subtract 4x from both sides and add the fives, so that: 2x = 10, so x = 5

3. Now check it:

6x = 6(5) = 30 for m<AOB

y + 5 = 4x +5 +5 = 4(5) + 10 = 30 for m<BOC. This makes sense because <AOB = <BOC.

Next, check the m<AOC. It needs to equal 60 since <AOB and <BOC are 30 each.

m<AO = 2X + 2y

2X = 2(5) = 10

2y = 2[4(5) + 5] = 2[20 + 5] = 2[25] = 50

Now add what you got for 2X to 2Y: 50 + 1- = 60.

Everything checks.

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