1 PROBLEM 1A PROBLEM 2A PROBLEM 1B PROBLEM 2B PROBLEM 3A PROBLEM 4BPROBLEM 4A PROBLEM 3B STANDARD 13...

1

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

http://www.mrperezonlinemathtutor.com/

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.

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved


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



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.



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:



7


A

CF

B

D

E

12

34

ADJACENT ANGLES:

Can you see other adjacent angles in the figure?

EFD DFCand ; common ray FD



8


A

CF

B

D

E

12

34

ADJACENT ANGLES:


CFB AFBand ; common ray FB



9


A

CF

B

D

E

12

34

ADJACENT ANGLES:


EFC AFCand ; common ray FC



10


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


A

CF

B

D

E

12

34

2 and 3 are ADJACENT ANGLES because they have a COMMON RAY FC

ADJACENT ANGLES:


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



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



13

STANDARD 13SUPPLEMENTARY ANGLES

RS

T

75°

What kind of angles are these two angles?

F

HG

105°



14



75°

F

HG

105°



15



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


64°

Find the measure of the supplement of the angle below:

180° – 64° = 116°

116°

Supplement:



17



180° –145° = 35°145°

Supplement:

35°



18


70°


180° – 70° = 110°

110°

Supplement:



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!



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



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



22

STANDARD 13COMPLEMENTARY ANGLES

ML

K

60°A

B

C30°




23


A

B

C30°

What kind of angle are these two angles?

60°



24



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


Find the measure of the COMPLEMENT of the angle below:

90° – 35° = 55°

Complement

55°

35°



26



90° – 67° = 23°67° 23°

Complement



27



90° – 39° = 51°

Complement

51°

39°



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



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



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.



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.


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


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


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:


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°


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:


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:


Date post:	23-Dec-2015
Category:	Documents
Upload:	augustine-sharp
View:	236 times
Download:	0 times

1 PROBLEM 1A PROBLEM 2A PROBLEM 1B PROBLEM 2B PROBLEM 3A PROBLEM 4BPROBLEM 4A PROBLEM 3B STANDARD 13...

Documents