Per: Homework 4: Angle Addition Postulate
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2
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1. Use the diagram below to complete each part.
a) Name the vertex of 24.
D
1
b) Name the sides of 21.
5 B
c) Write another name for 25.
E 4
d) Classify each angle:
F
ZFBC:
ZEBE:
ZABC:
g) Name an angle bisector.
• BF I AC
h) If mZEBD = 36° and mZDBC = 108°, find mZEBC.
i) If mZEBF = 117°, find mZABE.
2. If mZMKL = 83', mZJKL = 127", and
3. If mZEFH = (5x + 1), m_HFG = 62, and
mZJKM = 19% - 10)”, find the value of x.
mZEFG = (18x + 11), find each measure.
А
M1
5H

The measure of an angle, that forms a known larger angle with another

known angle can be determined by angle addition postulate.

Correct responses:

1. a) Point B

b) and

c) ∠EBD

d) ∠FBC = Right angle

e) ∠EBF = An obtuse angle

f) ∠ABC = Straight angle

g)  

h) m∠EBC = 180°

i) 36°

2)  x = 6°

3) x = 4°

a) The vertex of an angle is the point where the lines forming the angles meet.

b) The sides of an angle are the rays that form the angle.

c) The name of an angle can be given by the three points of the angle

Therefore;

d) Given that ⊥ , we have;

e) ∠EBF = An obtuse angle

f) ∠ABC = 180° = Straight angle

g) Given that by symbol for equal angles in the diagram, we have;

∠EBD = ∠ABE

Therefore, segment bisects ∠ABD

Which gives;

h) m∠EBD = 36°, m∠DBC = 108°

m∠EBC = m∠ABE + m∠EBD + m∠DBC  (angle addition property)

m∠EBC = m∠EBD + m∠EBD + m∠DBC (substitution property)

Therefore;

i) m∠EBF = 117°

m∠EBF = m∠ABE + m∠ABF

m∠ABF = m∠FBC = 90°

Therefore;

117° = m∠ABE + 90°

2. Given:

m∠MKL = 83°, m∠JKL = 127°, m∠JKM = (9·x - 10)°

Required:

The value of x

Solution:

m∠JKL = m∠MKL + m∠JKM

Which by plugging in the values gives;

127° = 83° + (9·x - 10)°

127° - 83° =  44° = (9·x - 10)°

44° + 10° = 54° = 9·x

3. m∠EFH = (5·x + 1)°

m∠HFG = 62°

m∠EFG = (18·x + 11)°

By angle addition property, we have;

m∠EFG = m∠EFH + m∠HFG

Therefore;

18·x + 11 = 5·x + 1 + 62

18·x - 5·x = 62 + 1 - 11 = 52

13·x = 52

Learn more about angle addition property here:

link

Answer/Step-by-step explanation:

a. Vertex of <4 is the endpoint where the 2 rays meet to form <4.

Vertex of <4 is point B.

b. Sides of <1 are rays BD and BC.

c. Another name for <5 is <DBE.

d. <FBC is a right angle. (BF meets AC at 90°)

e. <EBF is an obtuse angle. (It is greater than 90° but less than 180°)

f. <ABC is a straight angle. (It measures 180°)

g. An angle bisector is ray BE

h. m<EBD = 36°

m<DBC = 108°

m<EBD + m<DBC = m<EBC (angle addition postulate)

36 + 108 =  m<EBC (Substitution)

144° = m<EBC

m<EBC = 144°

i. m<EBF = 117°

m<ABE + m<ABF = m<EBF (angle addition postulate)

m<ABE + 90° = 117° (Substitution)

m<ABE = 117 - 90

m<ABE = 27°

2. m<MKL = 83°,

m<JKL = 127°,

m<JKM = (9x - 10)°

m<JKM + m<MKL = m<JKL (angle addition postulate)

(9x - 10)° + 83° = 127° (substitution)

Solve for x

9x - 10 + 83 = 127

9x + 73 = 127

9x + 73 - 73 = 127 - 73

9x = 54

9x/9 = 54/9

x = 6

3. m<EFH = (5x + 1)°

m<HFG = 62

m<EFG = (18x + 11)

m<EFH + m<HFG = m<EFG (angle addition postulate)

5x + 1 + 62 = 18x + 11 (substitution)

5x + 63 = 18x + 11

5x - 18x = -63 + 11

-13x = -52

-13x/-13 = -52/-13

x = 4

m<EFH = (5x + 1)°

Plug in the value of x

m<EFH = 5(4) + 1 = 21°

m<EFG = (18x + 11)

Plug in the value of x

m<EFG = 18(4) + 11 = 72 + 11

m<EFG = 83°

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