Brian ericson cep 810 11 16 09
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Angle Pairs. Brian Ericson CEP 810 11/16/09. Corresponding Angles. When lines are NOT II. When lines are II. Finding Corresponding Angles. Given : Line AB II Line CD, m ∠ ɑ=(5x+50) °, m ∠ β =(10x) ° Find : m ∠ ɑ and m ∠ β 5x+50=10x Subtract 5x 50 = 5x Divide by 5 X = 10

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Brian Ericson CEP 810 11/16/09

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Brian ericson cep 810 11 16 09

Angle Pairs

Brian Ericson

CEP 810

11/16/09


Corresponding angles

Corresponding Angles

When lines are NOT II

When lines are II


Finding corresponding angles

Finding Corresponding Angles

Given: Line AB II Line CD, m ∠ ɑ=(5x+50) °, m ∠β =(10x) °

Find: m ∠ ɑ and m ∠β

5x+50=10x

Subtract 5x

50 = 5x

Divide by 5

X = 10

m ∠ ɑ=(5(10)+50) °

m ∠ ɑ=100 °

m ∠β =(10(10)) °

m ∠β =100 °


Alternate interior angles

Alternate Interior Angles

When lines are NOT II

When lines are II


Finding alternate interior angles

Finding Alternate Interior Angles

Given: Line AB II Line CD, m ∠ ɑ=(4x) °, m ∠β =(2x+30) °

Find: m ∠ ɑ and m ∠β

2x+30=4x

Subtract 2x

30 = 2x

Divide by 2

15 = x

m ∠ ɑ=(4(15)) °

m ∠ ɑ=60 °

m ∠β =(2(15)+30) °

m ∠β =60 °


Alternate exterior angles

Alternate Exterior Angles

When lines are NOT II

When lines are II


Finding alternate exterior angles

Finding Alternate Exterior Angles

Given: Line AB II Line CD, m ∠ ɑ=(x+60) °, m ∠β =(3x) °

Find: m ∠ ɑ and m ∠β

x+60=3x

Subtract x

60 = 2x

Divide by 2

30 = x

m ∠ ɑ=((30)+60) °

m ∠ ɑ=90 °

m ∠β =(3(30)) °

m ∠β =90 °


Consecutive interior angles

Consecutive Interior Angles

When lines are NOT II

When lines are II


Finding consecutive interior angles

Finding Consecutive Interior Angles

Given: Line AB II Line CD, m ∠ ɑ=(40x+20) °, m ∠β =(30x+20) °

Find: m ∠ ɑ and m ∠β

40x+20+30x+20=180

Combine like terms

70x+40 =180

Subtract 40

70x =140

Divide by 70

X =2

m ∠ ɑ=(40(2)+20) °

m ∠ ɑ=100 °

m ∠β =(30(2)+20) °

m ∠β =80 °


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