Chapter 1 Section 6

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# Chapter 1 Section 6 - PowerPoint PPT Presentation

Chapter 1 Section 6. Exploring Angles. Warm-Up. 1) Name two segments whose midpoint is F in the figure below. D. E. F. G. H. 2) Find the coordinates of B is M(-3, 2) is the midpoint of line AB and A is at (-1, -1)

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## Chapter 1 Section 6

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### Chapter 1Section 6

Exploring Angles

Warm-Up
• 1) Name two segments whose midpoint is F in the figure below.

D

E

F

G

H

• 2) Find the coordinates of B is M(-3, 2) is the midpoint of line AB and A is at (-1, -1)
• 3) Line RS bisects line TU at point Q. Find the value of x and the measure of line QU if TU = 7x – 5 and TQ = -5x + 6

D

E

F

G

H

EG and DH

2) Find the coordinates of B, if M(-3, 2) is the midpoint of line AB and A is at (-1, -1).

• A and B are on a coordinate plane so use the equation
• [(x1 + x2)/2, (y1 + y2)/2].
• Let point A be x1 and y1 and let point B be x2 and y2.
• (x1 + x2)/2 = x-coordinate of the midpoint
• (-1 + x2)/2 = -3
• -1 + x2 = -6
• x2 = -5
• (y1 + y2)/2 = y-coordinate of the midpoint
• (-1+ y2)/2 = 2
• (-1+ y2) = 4
• y2 = 3
• So point B is at (-5,3).

3) Line RS bisects line TU at point Q. Find the value of x and the measure of line QU if TU = 7x – 5 and TQ = -5x + 6

• Since Q is the midpoint of line TU, we can use the midpoint formula. The midpoint formula tells us that TQ is congruent or equal to QU. So TQ + QU = TU; TQ + TQ = TU or 2(TQ) = TU.
• 2( TQ) = TU
• 2(-5x + 6) = 7x - 5
• -10x + 12= 7x - 5
• 12 = 17x – 5
• 17 = 17x
• 1 = x
• Plug 1 in for x in the equation for TQ.
• TQ = -5x + 6
• TQ = -5(1) + 6
• TQ = -5 + 6
• TQ = 1 = QU

T

Q

U

Vocabulary

Angle- A figure consisting of two noncollinear rays with a common endpoint.

Ray- PQ is a ray if it is the set of points consisting of PQ and all points S for which Q is between P and S.

Opposite rays- Two rays BA and BC such that B is between A and C.

Sides- The two rays that form the sides of the angle.

Vertex- The common endpoint of the two rays that form the angle.

P

Q

S

A

B

C

Side

Vertex

Side

Vocabulary Cont.

Interior – A point is in the interior of an angle if it does not lie on the angle itself and it lies on a segment whose endpoints are on the sides of the angle.

Exterior- A point that is not in the interior of the angle.

Degrees- A unit of measure used in measuring angles and arcs of circles.

Angle bisector- A ray that splits an angle into two congruent angles. Ray BD is an angle bisector.

Exterior

Interior

Exterior

A

D

B

C

Vocabulary Cont.

Right Angle-A 90 degree angle

Acute Angle-An angle less than 90 degrees

Obtuse Angle-An angle greater than 90 degrees

Straight Angle- An angle with a measure of 180 degrees.

Protractor Postulate- Given Ray AB and a number r between 0 and 180, there is exactly one ray with endpoint A, extending on either side of ray AB, such that the measure of the angle formed is r.

Right Angle

Acute Angle

Obtuse Angle

Straight Angle

Vocabulary Cont.

P

Angle Addition Postulate- If R is in the interior of angle PQS, then m<PQR + m<RQS = m<PQS. If m<PQR + m<RQS = m<PQS, then R is in the interior of angle PQS.

R

Q

S

A

D

1

2

B

3

C

• a) Give another name for angle ABD.
• <2 or <DBA
• b) Name the vertex of angle 3.
• Point C
• c) What are the sides of angle 1?
• Ray DA and ray DB

A

D

1

2

B

3

C

• a) Name a point in the interior of angle ABC.
• Point D
• b) Name a point in the exterior of angle CBD.
• Point A
• c) What is another name for angle ADB?
• <BDA or <1

Example 3: Refer to the figure below.

E

D

4

F

3

2

1

A

C

B

Does <EFC appear to be acute, right, obtuse, or straight?

Right

b) Complete: m<DFB = m<2 + ?

m<3

c) Name a point on the exterior of <AFB.

E, D, or C

Example 4: Suppose <T is congruent to <S, m<T = 12n – 6, and m<S = 4n + 18. Is <T acute, right, obtuse, or straight?

m<T = m<S

12n – 6 = 4n + 18

8n – 6 = 18

8n = 24

n = 3

Plug 3 in for n in the equation for m<T.

m<T = 12n – 6

m<T = 12(3) – 6

m<T = 36 - 6

m<T = 30

Since 30 is less than 90, angle T is an acute angle.

Example 5: In the figure, ray BA and ray BC are opposite rays, and ray BE bisects <ABD.

E

D

A

B

C

If m<ABE = 6x + 2 and m<DBE = 8x -14, find m<ABE.

Since ray BE bisects <ABD, we know that m<ABE = m<DBE

m<ABE = m<DBE

6x + 2 = 8x -14

2 = 2x – 14

16 = 2x

8 = x

Plug 8 in for x in the equation for m<ABE.

m<ABE = 6x + 2

m<ABE = 6(8) + 2

m<ABE = 48 + 2

m<ABE = 50

Example 6: In the figure, ray BA and ray BC are opposite rays, and ray BE bisects <ABD.

E

D

A

B

C

Given that m<ABD = 2y and m<DBC = 6y – 12, find m<DBC.

Since ray BA and ray BC are opposite rays, they form a straight angle.

m<ABD + m<DBC = 180

2y + 6y – 12 = 180

8y – 12 = 180

8y = 192

y = 24

Plug 24 in for y in the equation for m<DBC.

m<DBC = 6y - 12

m<DBC = 6(24) - 12

m<DBC = 144 - 12

m<DBC = 132

Example 7: In the figure, ray BA and ray BC are opposite rays, and ray BE bisects <ABD.

E

D

A

B

C

Find m<EBD if m<ABE = 12n – 8 and m<ABD = 22n - 11

Since ray BE bisects <ABD, m<ABE = m<EBD. Use the angle addition postulate m<ABE + m<EBD = m<ABD; 2(m<ABE) = m<ABD

2(m<ABE) = m<ABD

2(12n – 8) = 22n -11

24n – 16 = 22n -11

2n – 16 = -11

2n = 5

n = 5/2

Plug 5/2 in for n in the equation for m<ABE.

m<ABE = 12n - 8

m<ABE = 12(5/2) - 8

m<ABE = 30 - 8

m<ABE = 22 = m<EBD