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7. mAOB = 4x  1; mBOC = 2x + 15; mAOC = 8x + 8


8. mCOD = 8x + 13; mBOC = 3x  10; mBOD = 12x  6


9. ABC and EBF are a pair of vertical angles; mABC = 3x + 8 and mEBF = 2x + 48. What are mABC and mEBF?


10. JKL and MNP are complementary; mJKL = 2x  3 and mMNP = 5x + 2. What are mJKL and mMNP?

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Part 7.

From the diagram, m<AOC = m<AOB + m<BOC

8x + 8 = 4x - 1 + 2x + 15 = 6x + 14

8x - 6x = 14 - 8

2x = 6

x = 6 / 2 = 3

Therefore, m<AOC = 4(3) - 1 = 12 - 1 = 11

m<AOB = 2(3) + 15 = 6 + 15 = 21

m<BOC = 8(3) + 8 = 24 + 8 = 32



Part 8.

From the diagram, m<BOD = m<COD + m<BOC

12x - 6 = 8x + 13 + 3x - 10 = 11x + 3

12x - 11x = 3 + 6

x = 9

Therefore, m<COD = 8(9) + 13 = 72 + 13 = 85

m<BOC = 3(9) - 10 = 27 - 10 = 17

m<BOD = 12(9) - 6 = 108 - 6 = 102



Part 9.

Given that <ABC and <EBF are a pair of vertical angles and m<ABC = 3x + 8 and m<EBF = 2x + 48.

Recall that vertical angles are equal.
Thus 3x + 8 = 2x + 48
3x - 2x = 48 - 8 = 40
x = 40

Therefore, m<ABC = 3(40) + 8 = 120 + 8 = 128 and m<EBF = 2(40) + 48 = 80 + 48 = 128



Part 10.

Given that <JKL and <MNP are complementary and m<JKL = 2x - 3 and m<MNP = 5x + 2.

Recall that two angles are said to be complementary when their sum is 90 degrees.

Thus, 2x - 3 + 5x + 2 = 90
7x - 1 = 90
7x = 90 + 1 = 91
x = 91 / 7 = 13

Therefore, m<JKL = 2(13) - 3 = 26 - 3 = 23 and m<MNP = 5(13) + 2 = 65 + 2 = 67.

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