Question

-> OB bisects <AOC
m<AOB=6x, m<AOC=2x+2y, m<BOC=y+5

Answer 1

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.