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PROBLEM 1A
PROBLEM 2A
PROBLEM 1B
PROBLEM 2B
PROBLEM 3A
PROBLEM 4BPROBLEM 4A
PROBLEM 3B
STANDARD 13
ANGLES: INTRODUCTION
PROBLEM 5BPROBLEM 5A
ANGLES: CLASSIFICATION
ADJACENT ANGLES
ANGLE ADDITION POSTULATE
SUPPLEMENTARY ANGLES
LINEAR PAIR
COMPLEMENTARY ANGLES
VERTICAL ANGLES
END SHOWPRESENTATION CREATED BY SIMON PEREZ. All rights reserved
2
STANDARD 13:
Students prove relationships between angles in polygons using properties of complementary, supplementary, vertical and exterior angles.
ESTÁNDAR 13:
Los estudiantes prueban relaciones entre ángulos en polígonos usando propiedades de ángulos complementarios, suplementarios, verticales y ángulos exteriores.
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3
STANDARD 13What is an angle?
How many ways may we call this angle?
ABC
CBA
B
1AB
C
1
AB
C
Common endpoint called: VERTEX
Non-collinear ray
Non-collinear Ray
INTERIOR
EXTERIOR
EXTER
IOR
AB
C
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4
STANDARD 13 ANGLES
A
CF
B
D
E
12
34
How else may we call 2?
DFC or CFD
May be called F?
No, because point F has more than two rays departing from it.
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5
STANDARD 13ANGLES CLASSIFICATION
ACUTE ANGLE: OBTUSE ANGLE:RIGHT ANGLE:
0° < ANGLE < 90° ANGLE = 90° 90° < ANGLE < 180°
STRAIGHT ANGLE:
ANGLE = 180°PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
6
STANDARD 13ANGLE RELATIONSHIPS
A
CF
B
D
E
12
34
2 and 3 are ADJACENT ANGLES because they have a COMMON RAY FC
ADJACENT ANGLES:
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7
STANDARD 13ANGLE RELATIONSHIPS
A
CF
B
D
E
12
34
ADJACENT ANGLES:
Can you see other adjacent angles in the figure?
EFD DFCand ; common ray FD
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8
STANDARD 13ANGLE RELATIONSHIPS
A
CF
B
D
E
12
34
ADJACENT ANGLES:
Can you see other adjacent angles in the figure?
CFB AFBand ; common ray FB
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9
STANDARD 13ANGLE RELATIONSHIPS
A
CF
B
D
E
12
34
ADJACENT ANGLES:
Can you see other adjacent angles in the figure?
EFC AFCand ; common ray FC
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10
STANDARD 13ANGLE RELATIONSHIPS
A
CF
B
D
E
12
34
ADJACENT ANGLES:
Are and adjacent?EFD BFA
NO, because they don’t have a common ray.PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
11
STANDARD 13ANGLE RELATIONSHIPS
A
CF
B
D
E
12
34
2 and 3 are ADJACENT ANGLES because they have a COMMON RAY FC
ADJACENT ANGLES:
Can you see other adjacent angles in the figure?
EFD DFCand
CFB BFAand
EFC AFCand
Are and adjacent?EFD BFA
NO, because they don’t have a common ray.
; common ray FD
; common ray FB
; common ray FC
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12
STANDARD 13ANGLE RELATIONSHIPS: ANGLE ADDITION POSTULATE (Simplified)
LN
M
K
KNL m + LNMm = MNKm
Now may you find EFB?m
A
CF
B
D
E
EFB=m EFD m + DFCm + CFBm
It means: MEASURE of angle KNL
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13
STANDARD 13SUPPLEMENTARY ANGLES
RS
T
75°
What kind of angles are these two angles?
F
HG
105°
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STANDARD 13SUPPLEMENTARY ANGLES
What kind of angles are these two angles?
75°
F
HG
105°
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15
STANDARD 13SUPPLEMENTARY ANGLES
What kind of angles are these two angles?
TSR m + FGHm = 180°
Two angles that added together are 180°, are called: SUPPLEMENTARY ANGLES.
75°105°
TSR FGHAngles and are SUPPLEMENTARY!
F
HG
105°
RS
T
75°
and each one is the SUPPLEMENT of the other.PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
16
STANDARD 13SUPPLEMENTARY ANGLES
64°
Find the measure of the supplement of the angle below:
180° – 64° = 116°
116°
Supplement:
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17
STANDARD 13SUPPLEMENTARY ANGLES
Find the measure of the supplement of the angle below:
180° –145° = 35°145°
Supplement:
35°
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18
STANDARD 13SUPPLEMENTARY ANGLES
70°
Find the measure of the supplement of the angle below:
180° – 70° = 110°
110°
Supplement:
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19
STANDARD 13
105° 75°
Let’s take a closer look at the angle formation we got before:
• They are supplementary: 105° + 75° = 180°
• They are adjacent: have a common ray.
• They are LINEAR PAIR: because they are all of the above!
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20
32X + 10 8X +10 + = 180
40X +20 = 180-20 -20
40X = 16040 40
X= 4
This is a linear pair, so both angles are supplementary:
m RST = 32X + 10= 32( )+10
= 128 + 10
= 138°
m UST = 180-138
= 42°
4
R S U
T
32X + 10 8X + 10
STANDARD 13FIND: RST and UST
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21
20X + 10 5X +20 + = 18025X +30 = 180
-30 -30
25X = 15025 25
X= 6
This is a linear pair, so both angles are supplementary:
m KLN = 20X + 10= 20( )+10
= 120 + 10
= 130°
m MLN = 180-130
= 50°
6
K L M
N
20X + 10 5X + 20
STANDARD 13FIND: KLN and MLN
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22
STANDARD 13COMPLEMENTARY ANGLES
ML
K
60°A
B
C30°
What kind of angles are these two angles?
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23
STANDARD 13COMPLEMENTARY ANGLES
A
B
C30°
What kind of angle are these two angles?
60°
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24
STANDARD 13COMPLEMENTARY ANGLES
What kind of angles are these two angles?
60°
30°KLM m + ABCm = 90°
Two angles that added together are 90°, are called: COMPLEMENTARY ANGLES.
KLM ABCAngles and are complementary!
ML
K
60°
A
B
C30°
and each one is the COMPLEMENT of the other.PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
25
STANDARD 13SUPPLEMENTARY ANGLES
Find the measure of the COMPLEMENT of the angle below:
90° – 35° = 55°
Complement
55°
35°
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STANDARD 13COMPLEMENTARY ANGLES
Find the measure of the COMPLEMENT of the angle below:
90° – 67° = 23°67° 23°
Complement
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27
STANDARD 13COMPLEMENTARY ANGLES
Find the measure of the COMPLEMENT of the angle below:
90° – 39° = 51°
Complement
51°
39°
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28
9X + 10 20X-7 + = 90°29X +3 = 90
-3 -3
29X = 8729 29
Both angles are complementary:
m TRS = 9X+10= 9( ) + 10
= 27 + 10
m TRU = 90-37
3
X = 3
= 37°
= 53°
R
S
T
U
9X + 10
20X - 7
STANDARD 13FIND: TRS and TRU
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29
7X + 12 13X-2+ = 90°
20X+10 = 90-10 -10
20X = 8020 20
Both angles are complementary:
m FGT = 7X+12= 7( ) + 12
= 28+ 12
m HGT = 90-40
4
X = 4
= 40°
= 50°
G
F
T
H
7X + 12
13X - 2
STANDARD 13FIND: FGT and HGT
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30
STANDARD 13
q
s
Y
UV
X
Z
VERTICAL ANGLES
UZY VZX UZY=m VZXm
UZV YZX UZV=m YZXm
VERTICAL ANGLES are congruent, that is they have the same measure.
Observe they are straight lines
Means congruence
VERTICAL ANGLES are nonadjacent angles formed by two intersecting lines.
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31
STANDARD 13VERTICAL ANGLES
M
KV
X
L
Are and vertical angles?KLM VLX
No, although they look to be formed by two intersecting lines, in reality LK and LX are two rays that don’t lie in the same line and therefore the angles aren’t vertical.
Note: two red lines in one angle means that such angle is different to the other with just one red line.
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32
Both angles are vertical :
20 + 5X = 8X + 5
- 20 -20
5X = 8X -15
-8X -8X
-3X = -15-3 -3
X= 5
m ABC = 20 + 5X
= 20 + 5( )
m CBE = 180-45
5
= 45°
= 135°
m CBEm ABC and are linear pair
C
A
D
E
B20 + 5X 8X + 5
STANDARD 13FIND: ABC and CBE
= 20 + 25
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33
Both angles are vertical :
10 + 4X = 6X + 4
- 10 -10
4X = 6X -6
-6X -6X
-2X = -6-2 -2
X= 3
m SKR = 10 + 4X
= 10 + 4( )
m TKS = 180-22
3
= 22°
= 158°
m TKSm SKR and are linear pair
R
S
T
U
K10 + 4X 6X + 4
STANDARD 13FIND: SKR and TKS
= 10 + 12
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34
40°
STANDARD 13
m KJO + m NJO = 180°
m KJO + 40° = 180°- 40 - 40
m KJO = 140°
m MJN + m NJO = 90°
m MJN + 40° = 90°- 40 - 40
m MJN = 50°
m MJO = 90°
Two perpendicular lines form 4 right angles, so:
m NJO = m LJKIf by vertical angles:
m MJK = 90° + 40°
= 130°
140°
50°
then
130°
FIND:
K
LM
N
O
J
KJO, MJN, MJO MJKand
40°
By supplementary angles:
By complementary angles:
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35
STANDARD 13
m KJO + m NJO = 180°
m KJO + 35° = 180°- 35 - 35
m KJO = 145°
m MJN + m NJO = 90°
m MJN + 35° = 90°- 35 - 35
m MJN = 55°
m MJO = 90°
Two perpendicular lines form 4 right angles, so:
m NJO = m LJKIf by vertical angles:
m MJK = 90° + 35°
= 125°
145°
55°
then
125°
FIND:
K
LM
N
O
J
KJO, MJN, MJO MJKand
35°
By supplementary angles:
By complementary angles:
35°
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
36STANDARD 13
A
E
B
C
D
(12x + 10)°
(5x + 8)°
(3x + 2)°
Find andDEC, CEB, AEB, if rays ED and EA are opposite at point E.
DEC m AEBm++ CEBmDEA=m
DEA=m 180° by definition of a STRAIGHT ANGLE.
Now using all the above information and the figure:
(3x + 2) + (12x + 10) + (5x + 8) = 180°
3x + 12x + 5x + 2 + 10 + 8 = 180
20x + 20 = 180
-20 -20
20x = 16020 20
x = 8
Let’s find the measure of the angles:
m DEC = 3x + 2
= 3( ) + 28
= 24 + 2
= 26°
m CEB = 12x + 10
= 12( ) + 108
= 96 + 10
= 106°
m AEB = 5x + 8
= 5( ) + 88
= 40 + 8
= 48°
Verifying solution:
26° + 106° + 48° = 180°
180° = 180°
DEA,If rays ED and EA are opposite at point E then is a STRAIGHT ANGLE and the following holds true by Angle Addition Postulate:
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
37STANDARD 13
B
F
C
D
E
(13x + 12)°
(3x + 5)°
(4x + 3)°
Find andEFD, DFC, BFC, if rays FE and FB are opposite at point F.
EFD m BFCm++ DFCmEFB=m
EFB=m 180° by definition of a STRAIGHT ANGLE.
Now using all the above information and the figure:
(4x + 3) + (13x + 12) + (3x + 5) = 180°
4x + 13x + 3x + 3 + 12 + 5 = 180
20x + 20 = 180
-20 -20
20x = 16020 20
x = 8
Let’s find the measure of the angles:
m EFD = 4x + 3
= 4( ) + 38
= 32 + 3
= 35°
m DFC = 13x + 12
= 13( ) + 128
= 104 + 12
= 116°
m BFC = 3x + 5
= 3( ) + 58
= 24 + 5
= 29°Verifying solution:
35° + 116° + 29° = 180°
180° = 180°
EFB,If rays FE and FB are opposite at point F then is a STRAIGHT ANGLE and the following holds true by Angle Addition Postulate:
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