Algebra I by James Wenk © 2003 published by TEACHINGpoint as part of the Expert Systems for Teachers Series 1 3) 180 Rule for Triangles Objective-To find

Algebra I by James Wenk © 2003 published by TEACHINGpoint as part of the Expert Systems for Teachers Series

1

3) 180 Rule for Triangles

Objective-To find missing angles and to solve word problems involving geometric figures.

Some Strategies

1) Supplementary Angles

2) Complementary Angles

4) Vertical Angles

2:6 Word Problems Involving Geometric Figures


2

Supplementary Angles- Angles whose sum is 180 .

a b

ma + m b = 180

x 30

x + 30 = 180- 30 - 30

x = 150

Find the value of x.


3

Find the supplement of the given angle.

1) 40

2) 18

3) 153

4) 65

5) 89

6) 23

7) 131

8) 118

140

162

27

115

91

157

49

62


4

Write a variable equation and solve.

Find an angle whose supplement is 30 less than twice the angle.

x 2x - 30

x + (2x - 30) = 1803x - 30 = 180+30 +303x = 210

x = 7070


5

Complementary Angles - Angles whose sum is 90 .

ab

ma + m b = 90

x40

x + 40 = 90- 40 -40

x = 50


6

Find the complement of...

1) 20

2) 47

3) 100

4) the supplement of 150

70

43

No complement

the complement of the supplement of 150

30the complement of = 60


7

Write a variable equation and solve.

Find an angle whose complement is 20more than three times the angle.

x3x + 20

x + 3x + 20 = 90

4x + 20 = 90- 20 -204x = 704 4

x = 17.5


8

180 Rule for Triangles - the sum of the interiorangles of any triangle is always 180 .

a

b

c ma + m b + m c = 180

40

80

x

40 + 80 + x = 180120 + x = 180

x = 60


9

Find each angle below.

y +14 y - 20

2y - 10

(y + 14) + (2y - 10) + (y - 20) = 1804y - 16 = 180+16 +16

4y = 196y = 49

y + 14 = 63

2y - 10 = 88

y - 20 = 29

180


10

Vertical Angles Theorem - the opposite anglesof intersecting linesmust be equal.


11



12


Find the missing angles a, b and c.

25

a

bc

25

155


13


Find the missing angles a, b and c.

25

a

bc

25

155

155


14

Find the missing angle.

40 132

x

48

x = 180 - 40 - 48

x = 92


15

Find the missing angle.

119

x

12461

56

x = 180 - 61 - 56

x = 63


16

Use a variable equation to solve.1) The length of a rectangle is 5 less than 3 times its width. If the perimeter is 30 ft.,find its dimensions.

Let x = width3x - 5 = length

x

3x - 52(x) + 2(3x - 5) = 30

2x + 6x - 10 = 30+10 +10

8x = 408 8x = 5

= 5= 10


17

2) An angle is 6 degrees less than 3 times itscomplement. Find the angle.

Let x = the complement

x

3x - 6 = the angle

x + (3x - 6) = 90

(3x - 6)

4x - 6 = 90+6 +6

4x = 964 4x = 24

= 24

= 3(24) - 6 = 66


18

3) The largest angle in a triangle is four timesthe smallest. The third angle is 5 more than twice the smallest. Find each angle.

Let n = the smallest angle

2n + 5 = the middle angle

4n = the largest angle

n

2n + 54n

n + (2n + 5) + 4n = 1807n + 5 = 180

-5 -5

7n = 1757 7n = 25= 25

= 55

= 100


19

4) The lengths of the sides of a triangle areconsecutive even integers. If the perimeteris 24 inches, find the length of each side.

Let x = 1st sidex + 2 = 2nd sidex + 4 = 3rd side

x

x + 2

x + 4 x + (x + 2) + (x + 4) = 24

3x + 6 = 24-6 -63x = 183 3x = 6

= 6 in.= 8 in.= 10 in.

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Algebra I by James Wenk © 2003 published by TEACHINGpoint as part of the Expert Systems for Teachers Series 1 3) 180 Rule for Triangles Objective-To find