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PROBLEM 1PROBLEM 3

PROBLEM 2PROBLEM 4

PROBLEM 5

PROBLEM 8PROBLEM 7PROBLEM 6

STANDARD 13

SUPPLEMENT AND COMPLEMENT: NUMERIC

PROBLEM 10PROBLEM 9

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SUPPLEMENT: GENERAL

COMPLEMENT: GENERAL

SUPPLEMENT AND COMPLEMENT: GENERAL

WORD PROBLEMS:

SUPPLEMENTARY AND COMPLEMENTARY ANGLES

END SHOW

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STANDARD 7:

Students prove and use theorems involving the properties of parallellines cut by a transversal, the properties of quadrilaterals, andproperties of circles.

ESTÁNDAR 7:

Los estudiantes prueban y usan teoremas involucrando las propiedadesde líneas paralelas cortadas por una transversal, las propiedades de

cuadriláteros, y las propiedades de círculos.

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3

STANDARD 13

Find the measure of the COMPLEMENT and SUPPLEMENT of the angle below:

90° ± 65° = 25°65°

115°

Supplement:

180° ± 65° = 115°

Calculating the complement:

Calculating the supplement:

25°

Complement

65°

65°

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4

Find the measure of the COMPLEMENT and SUPPLEMENT of the angle below:

90° ± 55° = 35°55°

125°

Supplement:

180° ± 55° = 125°

Calculating the complement:

Calculating the supplement:

STANDARD 13

35°

Complement

55°

55°

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X°

Find the measure of the SUPPLEMENT of the angle below:

180° ± X°

(180 ± X)°

Supplement:

STANDARD 13

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6

Find the measure of the COMPLEMENT of the angle below:

90 ± X°

Complement

(90-X)°

X°

STANDARD 13

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Find COMPLEMENT AND SUPPLEMENT of any given angle:

90°± X°

(180 ± X)°

Supplement:

180° ± X°

Calculating the complement:

Calculating the supplement:

(90 ± X)°

Complement

X°

X°

Let¶s call our angle X:

X°

STANDARD 13

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8

X is the angle

180 ± X is the supplement

180 ± X =X ± 20

The supplement of an angle is twenty less than the angle, find this angleand the supplement.

+ X + X

180 = 2X ± 20+20 +20

200 = 2X

2 2X = 100 Then the supplement:

180 ± X = 180 ± 100

= 80

The angle is 100° and the supplement is 80°.

X180 ± X 

STANDARD 13

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9

X is the angle

90 ± X is the complement

90 ± X = 2X 

The complement of an angle is two times the angle, find this angle and thecomplement.

+ X + X

3 3

X = 30 Then the complement:

90 ± X = 90 ± 30= 60

The angle is 30° and the complement is 60°.

90 = 3X

X

90 - X

STANDARD 13

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10

X is the angle

90 ± X is the complement

90 ± X =X + 10

The complement of an angle is 10 more than the angle, find this angle andthe complement.

+X +X

90 = 2X + 10-10 -10

80 = 2X

2 2X =40 Then the complement:

90 ± X = 90 ± 40= 50

The angle is 40° and the complement is 50°.

X

90 - X

STANDARD 13

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11

X is the angle

180 ± X is the supplement

180 ± X =

The supplement of an angle is a third of the angle, find this angle and thesupplement.

540 = 4X4 4X = 135

Then the supplement:

180 ± X = 180 ± 135

= 45

The angle is 135° and the supplement is 45°.

X180 ± X 

X13

180 ± X = X13

33

540 ± 3X = X

+3X +3X

STANDARD 13

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X is the angle

180 ± X is the supplement

180 ± X =

The supplement of an angle is a fourth of the angle, find this angle and thesupplement.

720 = 5X5 5X = 144

Then the supplement:

180 ± X = 180 ± 144

= 36

The angle is 144° and the supplement is 36°.

X180 ± X 

X14

180 ± X = X14

44

720 ± 4X = X

+4X +4X

STANDARD 13

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13

X is the angle

90 ± X is the complement

90 ± X = 3X ± 10

The complement of an angle is 10 less than three times the angle, find thisangle and the complement.

+X +X

90 = 4X ± 10+10 +10

100 = 4X

4 4X = 25 Then the complement:

90 ± X = 90 ± 25= 65

The angle is 25° and the complement is 65°.

X

90 - X

STANDARD 13

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X is the angle

180 ± X is the supplement

180 ± X = 3X + 40

The supplement of an angle is forty more than the triple of the angle, findthis angle and the supplement.

+ X + X

180 = 4X + 40-40 -40

140 = 4X

4 4X = 35 Then the supplement:

180 ± X = 180 ± 35

= 145

The angle is 35° and the supplement is 145°.

X180 ± X 

STANDARD 13

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X is the angle

90 ± X is the complement

90 ± X = 2X ± 30

The complement of an angle is thirty less than twice the angle, find thisangle and the complement.

+X +X

90 = 3X ± 30+30 +30

120 = 3X

3 3X = 40 Then the complement:

90 ± X = 90 ± 40= 50

The angle is 40° and the complement is 50°.

X

90 - X

STANDARD 13

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X is the angle

180 ± X is the supplement

5(180 ± X) = (90 ± X) + 630

Five times the supplement of an angle is 630 more than its complement.Find the angle, the supplement and the complement.

+5X +5X

900 = 4X + 720-720 -720

180 = 4X4 4

X = 45

Then the supplement:

180 ± X = 180 ± 45

= 135

The angle is 45°, the supplement is 135° and the complement is 45°.

900 ± 5X = 720 ± X 

Then the complement:

90 ± X = 90 ± 45

= 45

STANDARD 13

90 ± X is the complementX180 ± X 

X

90 - X

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