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Honors Geometry

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

  • Honors Geometry -

    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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