Student's completed worksheet on the Angle Addition Postulate with geometric diagrams and calculations.

Let's go through the homework step by step and verify or correct each part of the problems. The image shows a geometry worksheet focused on angle addition postulate, angle classification, and solving for unknown angles using algebra.

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Problem 1: Use the diagram to complete each part

Given:
- Diagram with point $ B $ at center, rays going to points $ A, C, D, E, F $
- $ \overrightarrow{BF} \perp \overrightarrow{AC} $ → So $ \angle FBC = 90^\circ $
- $ m\angle DBC = 108^\circ $
- $ m\angle EBD = 36^\circ $

#### a) Name the vertex of $ \angle 4 $.
✔ Answer: B
✔️ Correct. The vertex is the common endpoint of the two rays forming the angle. For $ \angle 4 $, it’s at point $ B $.

#### b) Name the sides of $ \angle 1 $.
✔ Answer: BE, BD
✔️ Correct. $ \angle 1 $ is $ \angle EBD $, so its sides are rays $ \overrightarrow{BE} $ and $ \overrightarrow{BD} $. (Note: The student wrote "BE", but should include both sides.)

#### c) Write another name for $ \angle 5 $.
✔ Answer: $ \angle CBD $
Wait — the student wrote $ \angle EBD $, which is incorrect.

But $ \angle 5 $ is between $ \overrightarrow{BD} $ and $ \overrightarrow{BE} $? Wait — looking at the diagram:

From the labels:
- $ \angle 1 $: $ \angle EBD $
- $ \angle 2 $: $ \angle DBC $
- $ \angle 3 $: $ \angle CBF $
- $ \angle 4 $: $ \angle FBA $
- $ \angle 5 $: $ \angle ABE $

Wait — let's recheck the labeling.

Looking at the diagram:

- Ray $ \overrightarrow{BD} $ splits $ \angle ABC $ into $ \angle 1 $ and $ \angle 2 $
- $ \angle 1 $: $ \angle EBD $
- $ \angle 2 $: $ \angle DBC $
- $ \angle 3 $: $ \angle CBF $
- $ \angle 4 $: $ \angle FBA $
- $ \angle 5 $: $ \angle ABE $

So $ \angle 5 $ is formed by rays $ \overrightarrow{BA} $ and $ \overrightarrow{BE} $ → so another name is $ \angle ABE $

✘ Student wrote: $ \angle EBD $ → that's $ \angle 1 $, not $ \angle 5 $

✔ Correct answer: $ \angle ABE $

#### d) Classify each angle:
- $ \angle FBC $: $ 90^\circ $ → right angle ✔
- $ \angle EBF $: $ 90^\circ $ → right angle ✔
- $ \angle ABC $: $ 180^\circ $ → straight angle ✔

All correct.

#### g) Name an angle bisector.
Student wrote: $ \overrightarrow{DBF} $ ✘

An angle bisector divides an angle into two equal parts.

From the diagram:
- $ \angle DBC = 108^\circ $
- $ \angle EBD = 36^\circ $
- But $ \angle EBC = \angle EBD + \angle DBC = 36^\circ + 108^\circ = 144^\circ $

No indication that any ray bisects an angle unless specified.

But wait — in part h), we’re told $ m\angle EBD = 36^\circ $, $ m\angle DBC = 108^\circ $, so $ \angle EBC = 144^\circ $

Is $ \overrightarrow{BD} $ bisecting $ \angle EBC $? No — because $ 36^\circ \ne 108^\circ $

But maybe $ \overrightarrow{BF} $ is perpendicular to $ AC $, and $ \angle FBC = 90^\circ $

We don’t have enough info to say $ \overrightarrow{BD} $ or $ \overrightarrow{BF} $ is a bisector unless angles are equal.

Wait — perhaps $ \overrightarrow{BD} $ is not a bisector.

But the student wrote $ \overrightarrow{DBF} $, which is not even a ray — it’s a line segment from D to B to F.

A bisector must be a ray starting at the vertex.

Possibly, the student meant $ \overrightarrow{BD} $ or $ \overrightarrow{BF} $?

But no clear evidence that either bisects an angle.

Wait — maybe $ \angle ABC $ is straight ($ 180^\circ $), and $ \overrightarrow{BF} \perp \overrightarrow{AC} $, so $ \angle FBC = 90^\circ $, $ \angle FBA = 90^\circ $

So $ \overrightarrow{BF} $ splits $ \angle ABC $ into two $ 90^\circ $ angles.

✔ Therefore, $ \overrightarrow{BF} $ is the angle bisector of $ \angle ABC $

So the correct answer is: $ \overrightarrow{BF} $

✘ Student wrote $ \overrightarrow{DBF} $ — invalid notation; likely meant $ \overrightarrow{BF} $

✔ Correct: $ \overrightarrow{BF} $

#### h) If $ m\angle EBD = 36^\circ $ and $ m\angle DBC = 108^\circ $, find $ m\angle EBC $

$ \angle EBC = \angle EBD + \angle DBC = 36^\circ + 108^\circ = 144^\circ $

✘ Student wrote: $ 36^\circ $ → Incorrect

✔ Correct: $ 144^\circ $

#### i) If $ m\angle EBF = 117^\circ $, find $ m\angle ABE $

We know:
- $ \angle EBF = 117^\circ $
- $ \angle FBA = ? $
- $ \angle ABC = 180^\circ $ (straight angle)
- $ \angle EBF $ and $ \angle ABE $ are adjacent angles that together make $ \angle ABF $, but we need to see how they fit.

Wait — let's look carefully.

Points: $ A, B, C $ lie on a straight line (since $ \angle ABC = 180^\circ $)

$ \overrightarrow{BF} \perp \overrightarrow{AC} $ → so $ \angle FBC = 90^\circ $, $ \angle FBA = 90^\circ $

So $ \angle ABF = 90^\circ $

Now, $ \angle EBF = 117^\circ $? But $ \angle ABF = 90^\circ $, and $ \angle EBF $ is part of $ \angle ABF $? That can't be — $ 117^\circ > 90^\circ $

Contradiction.

Wait — maybe the diagram shows $ \angle EBF $ including $ \angle EBD $, $ \angle DBC $, and $ \angle CBF $? Let's reconstruct.

From the diagram:

- $ \angle EBF $ is composed of $ \angle EBD + \angle DBC + \angle CBF $? But $ \angle DBC = 108^\circ $, $ \angle CBF = 90^\circ $, already over $ 180^\circ $

Wait — something's wrong.

Wait — perhaps $ \angle EBF $ is the angle from $ E $ to $ F $ passing through $ B $

But $ \angle EBD = 36^\circ $, $ \angle DBC = 108^\circ $, so $ \angle EBC = 144^\circ $

Then $ \angle CBF = 90^\circ $, so total $ \angle EBF = \angle EBC + \angle CBF = 144^\circ + 90^\circ = 234^\circ $? That doesn't make sense.

Wait — angles around a point sum to $ 360^\circ $

Let’s assume:

- $ \angle ABC = 180^\circ $ (straight line)
- $ \overrightarrow{BF} \perp \overrightarrow{AC} $ → $ \angle FBC = 90^\circ $, $ \angle FBA = 90^\circ $
- $ \angle EBD = 36^\circ $
- $ \angle DBC = 108^\circ $
- So $ \angle EBD + \angle DBC = 36^\circ + 108^\circ = 144^\circ $ → this is $ \angle EBC $
- Then $ \angle EBF = \angle EBC + \angle CBF = 144^\circ + 90^\circ = 234^\circ $? But student says $ m\angle EBF = 117^\circ $

That suggests $ \angle EBF $ is not going through $ C $

Perhaps $ \angle EBF $ is the smaller angle from $ E $ to $ F $, passing through $ D $ and $ C $? But then it would be large.

Alternatively, maybe $ \angle EBF $ is measured clockwise or counterclockwise.

Wait — perhaps $ \angle EBF $ includes $ \angle EBD $, $ \angle DBF $? But $ \angle DBF $ is not labeled.

Wait — perhaps $ \angle EBF $ is made up of $ \angle EBD $ and $ \angle DBF $

But we don’t know $ \angle DBF $

But we do know $ \angle DBC = 108^\circ $, $ \angle CBF = 90^\circ $, so $ \angle DBF = \angle DBC + \angle CBF = 108^\circ + 90^\circ = 198^\circ $? Too big.

This is confusing.

Wait — maybe $ \angle EBF $ is the angle from $ E $ to $ F $ via the other side.

But $ \angle EBF = 117^\circ $, and we are to find $ \angle ABE $

Let’s suppose $ \angle ABE $ is the angle from $ A $ to $ E $, and $ \angle EBF $ is from $ E $ to $ F $, and together they might make $ \angle ABF = 90^\circ $

But $ \angle EBF = 117^\circ > 90^\circ $, so impossible.

Unless $ \angle EBF $ is not inside $ \angle ABF $

Wait — perhaps $ \angle EBF $ goes from $ E $ to $ F $ passing through $ D $, $ C $, etc.

But $ \angle EBD = 36^\circ $, $ \angle DBC = 108^\circ $, $ \angle CBF = 90^\circ $, so $ \angle EBF = 36^\circ + 108^\circ + 90^\circ = 234^\circ $

But student says $ m\angle EBF = 117^\circ $ — so likely, $ \angle EBF $ is the other way around.

Wait — angles can be measured in two directions.

So if $ \angle EBF = 117^\circ $, and $ \angle ABC = 180^\circ $, $ \angle FBC = 90^\circ $, $ \angle FBA = 90^\circ $

Then $ \angle EBF $ could be measured from $ E $ to $ F $ passing through $ A $

So $ \angle EBF = \angle EBA + \angle ABF $

But $ \angle ABF = 90^\circ $, so $ \angle EBA = \angle EBF - \angle ABF = 117^\circ - 90^\circ = 27^\circ $

So $ m\angle ABE = 27^\circ $

But student wrote $ 36^\circ $

Wait — maybe $ \angle EBF $ includes $ \angle EBD $ and $ \angle DBF $, but we need to use known values.

Let me try a different approach.

Assume:

- $ \angle EBD = 36^\circ $
- $ \angle DBC = 108^\circ $
- $ \angle CBF = 90^\circ $
- $ \angle FBA = 90^\circ $

Then total around point B: $ 36^\circ + 108^\circ + 90^\circ + 90^\circ = 324^\circ $ — missing $ 36^\circ $, which must be $ \angle ABE $? But $ \angle ABE $ is already included in $ \angle FBA $

Wait — perhaps the rays are ordered: $ \overrightarrow{BA}, \overrightarrow{BE}, \overrightarrow{BD}, \overrightarrow{BC}, \overrightarrow{BF} $

So angles:

- $ \angle ABE = x $
- $ \angle EBD = 36^\circ $
- $ \angle DBC = 108^\circ $
- $ \angle CBF = 90^\circ $
- $ \angle FBA = 90^\circ $ — but this overlaps

Wait — $ \angle FBA $ is from $ F $ to $ A $, which may include $ \angle FBC $, $ \angle CBA $ — no.

Actually, $ \angle FBA $ is from $ F $ to $ A $, and since $ \overrightarrow{BF} \perp \overrightarrow{AC} $, and $ A-B-C $ straight, then $ \angle FBA = 90^\circ $, $ \angle FBC = 90^\circ $

So $ \angle ABF = 90^\circ $

Now, $ \angle EBF = 117^\circ $ — but $ \angle ABF = 90^\circ $, so $ \angle EBF $ cannot be larger than $ \angle ABF $ if $ E $ is between $ A $ and $ F $

Unless $ E $ is on the other side.

Wait — perhaps $ \angle EBF $ is measured across the top.

But $ \angle EBD = 36^\circ $, so $ \angle EBA $ is part of $ \angle ABE $

Let’s assume the order of rays from $ B $ is:

Clockwise: $ \overrightarrow{BA}, \overrightarrow{BE}, \overrightarrow{BD}, \overrightarrow{BC}, \overrightarrow{BF} $

Then:
- $ \angle ABE = ? $
- $ \angle EBD = 36^\circ $
- $ \angle DBC = 108^\circ $
- $ \angle CBF = 90^\circ $
- $ \angle FBA = 90^\circ $ — but $ \angle FBA $ is from $ F $ to $ A $, which would be $ \angle FBC + \angle CBA $, but $ \angle CBA = 180^\circ $, so too big.

Wait — better: since $ A-B-C $ is straight, $ \angle ABC = 180^\circ $

$ \overrightarrow{BF} \perp \overrightarrow{AC} $, so $ \angle FBC = 90^\circ $, $ \angle FBA = 90^\circ $

So $ \angle ABF = 90^\circ $, $ \angle FBC = 90^\circ $

Now, $ \angle EBF $ is the angle from $ E $ to $ F $, which could be $ \angle EBA + \angle ABF $ if $ E $ is on the same side as $ A $

But $ \angle ABF = 90^\circ $, so $ \angle EBF = \angle EBA + 90^\circ $

If $ m\angle EBF = 117^\circ $, then $ \angle EBA = 117^\circ - 90^\circ = 27^\circ $

So $ m\angle ABE = 27^\circ $

But student wrote $ 36^\circ $ — incorrect.

✔ Correct: $ 27^\circ $

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Problem 2:

If $ m\angle MKL = 83^\circ $, $ m\angle JKL = 127^\circ $, and $ m\angle JKM = (9x - 10)^\circ $, find $ x $

From the diagram, $ \angle JKL = \angle JKM + \angle MKL $

So:
$$
m\angle JKL = m\angle JKM + m\angle MKL
$$
$$
127^\circ = (9x - 10)^\circ + 83^\circ
$$
$$
127 = 9x - 10 + 83
$$
$$
127 = 9x + 73
$$
$$
127 - 73 = 9x
$$
$$
54 = 9x
$$
$$
x = 6
$$

✔ Student got $ x = 6 $, correct.

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Problem 3:

If $ m\angle EFH = (5x + 1)^\circ $, $ m\angle HFG = 62^\circ $, and $ m\angle EFG = (18x + 11)^\circ $, find each measure.

From diagram: $ \angle EFG = \angle EFH + \angle HFG $

So:
$$
(18x + 11) = (5x + 1) + 62
$$
$$
18x + 11 = 5x + 63
$$
$$
18x - 5x = 63 - 11
$$
$$
13x = 52
$$
$$
x = 4
$$

Now:
- $ m\angle EFH = 5(4) + 1 = 20 + 1 = 21^\circ $
- $ m\angle HFG = 62^\circ $
- $ m\angle EFG = 18(4) + 11 = 72 + 11 = 83^\circ $

Check: $ 21 + 62 = 83 $ — correct.

Student had:
- $ 5x + 1 + 62 = 18x + 11 $
- $ 62 = 13x + 11 $ → mistake!
- $ 51 = 13x $ → $ x = 51/13 \approx 3.92 $ — incorrect

✔ Correct: $ x = 4 $, $ m\angle EFH = 21^\circ $, $ m\angle EFG = 83^\circ $

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Problem 4:

If $ m\angle MNP = (3x - 5)^\circ $, $ m\angle MNO = (9x - 15)^\circ $, and $ m\angle ONP = 30^\circ $, find measures.

Diagram shows $ \angle MNO = \angle MNP + \angle PNO $? Wait — likely $ \angle MNO = \angle MNP + \angle PNO $

But $ \angle ONP = 30^\circ $, so $ \angle PNO = 30^\circ $

So:
$$
m\angle MNO = m\angle MNP + m\angle PNO
$$
$$
(9x - 15) = (3x - 5) + 30
$$
$$
9x - 15 = 3x + 25
$$
$$
9x - 3x = 25 + 15
$$
$$
6x = 40
$$
$$
x = \frac{40}{6} = \frac{20}{3} \approx 6.67
$$

Then:
- $ m\angle MNP = 3(\frac{20}{3}) - 5 = 20 - 5 = 15^\circ $
- $ m\angle MNO = 9(\frac{20}{3}) - 15 = 60 - 15 = 45^\circ $
- $ m\angle ONP = 30^\circ $

Check: $ 15 + 30 = 45 $ — correct.

Student had:
- $ 9x - 15 = 3x - 5 + 30 $
- $ 9x - 15 = 3x + 25 $
- $ 6x = 40 $ → $ x = 40/6 = 20/3 $
- Then $ m\angle MNP = 3(20/3) - 5 = 20 - 5 = 15^\circ $
- $ m\angle MNO = 9(20/3) - 15 = 60 - 15 = 45^\circ $

✔ Student did this correctly.

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Problem 5:

If $ m\angle MOP = (5x + 3)^\circ $, $ m\angle NOP = (3x - 4)^\circ $, and $ \angle MON $ is a right angle, find measures.

Diagram: $ \angle MON = \angle MOP + \angle PON $? Wait — likely $ \angle MON = \angle MOP + \angle PON $

But $ \angle MON = 90^\circ $

And $ \angle NOP = (3x - 4)^\circ $, so $ \angle PON = 3x - 4 $

So:
$$
m\angle MON = m\angle MOP + m\angle PON
$$
$$
90 = (5x + 3) + (3x - 4)
$$
$$
90 = 8x - 1
$$
$$
91 = 8x
$$
$$
x = \frac{91}{8} = 11.375
$$

Then:
- $ m\angle MOP = 5(91/8) + 3 = 455/8 + 24/8 = 479/8 = 59.875^\circ $
- $ m\angle NOP = 3(91/8) - 4 = 273/8 - 32/8 = 241/8 = 30.125^\circ $
- Sum: $ 59.875 + 30.125 = 90^\circ $ — correct.

But student wrote:
- $ 5x + 3 + 3x - 4 = 90 $
- $ 8x - 1 = 90 $
- $ 8x = 91 $
- $ x = 91/8 $
- Then $ m\angle MOP = 5(91/8) + 3 = 455/8 + 24/8 = 479/8 = 59.875^\circ $
- $ m\angle NOP = 3(91/8) - 4 = 273/8 - 32/8 = 241/8 = 30.125^\circ $

✔ Student did this correctly.

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✔ Summary of Corrections:

| Problem | Part | Student Answer | Correct Answer |
|--------|------|----------------|----------------|
| 1a | Vertex of ∠4 | B | ✔ Correct |
| 1b | Sides of ∠1 | BE | Should be $ \overrightarrow{BE}, \overrightarrow{BD} $ |
| 1c | Another name for ∠5 | ∠EBD | ✘ Should be $ \angle ABE $ |
| 1d | Classify angles | All correct | ✔ |
| 1g | Angle bisector | $ \overrightarrow{DBF} $ | ✘ Should be $ \overrightarrow{BF} $ |
| 1h | $ m\angle EBC $ | 36° | ✘ Should be $ 144^\circ $ |
| 1i | $ m\angle ABE $ | 36° | ✘ Should be $ 27^\circ $ |
| 2 | $ x $ | 6 | ✔ Correct |
| 3 | $ x $ | 51/13 | ✘ Should be $ x=4 $, $ \angle EFH=21^\circ $, $ \angle EFG=83^\circ $ |
| 4 | $ x=20/3 $, angles | ✔ Correct |
| 5 | $ x=91/8 $, angles | ✔ Correct |

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Final Answers:

#### Problem 1:
a) B
b) $ \overrightarrow{BE}, \overrightarrow{BD} $
c) $ \angle ABE $
d) $ \angle FBC = 90^\circ $ (right), $ \angle EBF = 90^\circ $ (right), $ \angle ABC = 180^\circ $ (straight)
g) $ \overrightarrow{BF} $
h) $ 144^\circ $
i) $ 27^\circ $

#### Problem 2: $ x = 6 $

#### Problem 3: $ x = 4 $, $ m\angle EFH = 21^\circ $, $ m\angle EFG = 83^\circ $

#### Problem 4: $ x = \frac{20}{3} $, $ m\angle MNP = 15^\circ $, $ m\angle MNO = 45^\circ $

#### Problem 5: $ x = \frac{91}{8} $, $ m\angle MOP = \frac{479}{8}^\circ = 59.875^\circ $, $ m\angle NOP = \frac{241}{8}^\circ = 30.125^\circ $

Let me know if you'd like a cleaned-up version!