Student's completed geometry worksheet demonstrating how to use the Angle Addition Postulate to find angle measures.

Let's go through the homework step by step and verify or correct each part of the problems. The worksheet is about angle addition postulate and basic geometry concepts.

---

Problem 1: Use the diagram below to complete each part

#### Diagram Notes:
- Point B is the vertex of several angles.
- Rays: BA, BE, BD, BC, BF
- Given: $ \overrightarrow{BF} \perp \overrightarrow{AC} $ → So angle between BF and AC is 90°
- $ m\angle DBC = 108^\circ $
- $ m\angle EBD = 36^\circ $

---

#### a) Name the vertex of ∠4

- ∠4 is formed by rays BE and BA, so the vertex is B
✔ Correct: B

---

#### b) Name the sides of ∠1

- ∠1 is formed by rays BD and BC → sides are BD and BC
✘ Incorrect: You wrote "BE" — that’s not a side of ∠1.

✔ Correct answer: BD and BC

---

#### c) Write another name for ∠5

- ∠5 is formed by rays BE and BD → it's the same as ∠EBD
✔ Correct: ∠EBD

---

#### d) Classify each angle:

- ∠FBC: From diagram, BF ⊥ AC, and since AC is horizontal and BF is vertical, ∠FBC is 90°, so it's a right angle
✔ Correct: 90° (but classification should be “right angle”)

- ∠EBF: From diagram, this is formed by BE and BF. Since BF is perpendicular to AC, and BE is going toward A, we can see that ∠EBF is 90° (because BE is along AC, and BF is perpendicular). So it's also a right angle
✔ Correct: 90°

- ∠ABC: This is a straight line from A to C through B → it's a straight angle, so 180°
✔ Correct: 180°

So classification:
- ∠FBC: Right angle
- ∠EBF: Right angle
- ∠ABC: Straight angle

---

#### g) Name an angle bisector

- An angle bisector divides an angle into two equal parts.
- We’re told $ m\angle EBD = 36^\circ $, $ m\angle DBC = 108^\circ $
- Then $ m\angle EBC = m\angle EBD + m\angle DBC = 36 + 108 = 144^\circ $? Wait — no!

Wait: Actually, look at the diagram. Points: D, B, C — with ∠DBC = 108°, and ∠EBD = 36°, but they may not be adjacent in a way that adds directly.

But wait — if you look at the diagram carefully:

- ∠EBD = 36°
- ∠DBC = 108°
- And these are adjacent angles sharing ray BD.

So total ∠EBC = ∠EBD + ∠DBC = 36° + 108° = 144°?

But then later, part h says: If ∠EBD = 36°, ∠DBC = 108°, find ∠EBC → you wrote 36°, which is wrong.

Wait — let's re-express.

Hold on: There might be a misunderstanding in labeling.

But looking at your work:

You said:
> h) If m∠EBD = 36° and m∠DBC = 108°, find m∠EBC: 36° ← ✘

That’s incorrect.

Actually, ∠EBC is composed of ∠EBD and ∠DBC only if D lies between E and C.

From the diagram:
- Ray BE, BD, BC
- So if D is between E and C, then ∠EBC = ∠EBD + ∠DBC = 36° + 108° = 144°

But you wrote 36° — that’s wrong.

Also, in part i): If m∠EBF = 117°, find m∠ABE

We know:
- BF ⊥ AC → so ∠ABF = 90°
- But ∠EBF = 117°
- So ∠ABE = ∠EBF - ∠ABF = 117° - 90° = 27°

But you wrote 36° — again, incorrect.

Let’s fix everything.

---

✔ Correcting Problem 1:

#### a) Vertex of ∠4 → B ✔
#### b) Sides of ∠1 → BD and BC ✘ (you said BE)
#### c) Another name for ∠5 → ∠EBD ✔
#### d) Classify:
- ∠FBC: Right angle (90°) ✔
- ∠EBF: Right angle (90°) ✔
- ∠ABC: Straight angle (180°) ✔

#### g) Angle bisector → You wrote DBF — but DBF is not a ray.

Wait — what is an angle bisector?

Look: ∠EBD = 36°, ∠DBC = 108° — is there a ray that splits one of these?

But we don’t have info about bisectors.

Wait — maybe BD is the bisector of some angle?

But ∠EBC = 36° + 108° = 144°, and BD splits it into 36° and 108° — not equal → not bisector.

Alternatively, perhaps BF? But no.

Wait — maybe BE? No.

No obvious bisector unless more info.

But maybe BD is bisecting ∠EBC? No — 36 ≠ 108.

Wait — perhaps the question wants us to identify a bisector based on given info.

But nothing indicates a bisector.

Wait — perhaps BF is the bisector of ∠ABC?

But ∠ABC is 180°, and BF is perpendicular to AC → so it splits it into two 90° angles → yes!

So BF is the angle bisector of ∠ABC because it splits the straight angle into two 90° angles.

So correct answer: BF

You wrote DBF — that’s not a ray. Probably meant BF

✔ So g) Answer: BF

---

#### h) If m∠EBD = 36° and m∠DBC = 108°, find m∠EBC

Since both angles share ray BD and are adjacent, and E-B-D-C are in order around point B:

So ∠EBC = ∠EBD + ∠DBC = 36° + 108° = 144°

✘ You wrote 36° — incorrect.

✔ Correct: 144°

---

#### i) If m∠EBF = 117°, find m∠ABE

We know:
- BF ⊥ AC → so ∠ABF = 90°
- ∠EBF = 117°
- ∠ABE = ∠EBF - ∠ABF = 117° - 90° = 27°

Because ray BA is along AC, and BE is between BA and BF.

So ∠ABE = 27°

✘ You wrote 36° — incorrect.

✔ Correct: 27°

---

Summary for Problem 1:

| Part | Your Answer | Correct Answer |
|------|-------------|----------------|
| a) | B | ✔ B |
| b) | BE | ✘ BD and BC |
| c) | ∠EBD | ✔ ∠EBD |
| d) | 90°, 90°, 180° | ✔ |
| g) | DBF | ✘ BF |
| h) | 36° | ✘ 144° |
| i) | 36° | ✘ 27° |

---

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: Point K, rays KM, KL, KJ

Angles:
- ∠MKL = 83°
- ∠JKL = 127°
- ∠JKM = 9x - 10

Note: ∠JKL = ∠JKM + ∠MKL

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

✔ You wrote:
```
9x - 10 = 44
9x = 54
x = 6
```

But where did 44 come from?

Wait — you probably did:
127 - 83 = 44 → so ∠JKM = 44° → then 9x - 10 = 44 → x = 6

Yes, that’s correct.

So even though you didn’t write the full equation, your logic is right.

✔ x = 6 → Correct

---

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: E-H-G-F, so ∠EFG = ∠EFH + ∠HFG

So:
$$
\angle EFG = \angle EFH + \angle HFG \\
18x + 11 = (5x + 1) + 62 \\
18x + 11 = 5x + 63 \\
18x - 5x = 63 - 11 \\
13x = 52 \\
x = 4
$$

Now plug back:
- $ 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 → ✔

You wrote:
```
5x+1 +62 = 18x+11
63 = 13x
x=?
```

Then:
- x = 4
- m∠EFH = 21°
- m∠EFG = 83°

✔ All correct.

---

Problem 4:

> If $ m\angle MOP = (3x + 3)^\circ $, $ m\angle NOP = (9x - 15)^\circ $, and $ m\angle MON = 63^\circ $, find each measure.

Diagram: M-O-N-P, so ∠MON = ∠MOP + ∠PON? Wait — need to check.

Wait: ∠MON = 63°, ∠MOP = 3x+3, ∠NOP = 9x-15

But if O is the vertex, and points M, N, P, likely: M-O-N, N-O-P → so ∠MON and ∠NOP are adjacent.

But ∠MOP = ∠MON + ∠NOP?

Yes, if N is between M and P.

So:
$$
\angle MOP = \angle MON + \angle NOP \\
3x + 3 = 63 + (9x - 15) \\
3x + 3 = 63 + 9x - 15 \\
3x + 3 = 9x + 48 \\
3 - 48 = 9x - 3x \\
-45 = 6x \\
x = -7.5
$$

Wait — negative angle? That can't be.

But you wrote:
```
3x + 3 + 9x - 15 = 63
12x - 12 = 63
12x = 75
x = 6.25
```

Wait — you assumed ∠MOP + ∠NOP = ∠MON? That doesn't make sense.

Wait — perhaps the diagram shows:

- M-O-N: ∠MON = 63°
- N-O-P: ∠NOP = 9x - 15
- M-O-P: ∠MOP = 3x + 3

But if M-O-N-P, then ∠MOP = ∠MON + ∠NOP

So:
$$
\angle MOP = \angle MON + \angle NOP \\
3x + 3 = 63 + (9x - 15) \\
3x + 3 = 9x + 48 \\
3 - 48 = 9x - 3x \\
-45 = 6x \\
x = -7.5
$$

Still negative.

But angles can't be negative.

Wait — maybe ∠MOP = 3x + 3 is the larger one?

Wait — perhaps ∠MOP = ∠MON + ∠NOP → so 3x+3 = 63 + (9x - 15)

But that gives x = -7.5 → impossible.

Alternative: Maybe ∠MOP is not the sum?

Wait — perhaps the diagram has ray ON between OM and OP.

So ∠MOP = ∠MON + ∠NOP

So:
$$
3x + 3 = 63 + (9x - 15) \\
3x + 3 = 9x + 48 \\
-6x = 45 \\
x = -7.5
$$

Still invalid.

But you wrote:
```
3x + 3 + 9x - 15 = 63
12x - 12 = 63
12x = 75
x = 6.25
```

This implies:
∠MOP + ∠NOP = ∠MON → which would mean they overlap or something.

But that doesn't make sense.

Wait — perhaps ∠MOP and ∠NOP are adjacent, and ∠MON is the difference?

Wait — maybe the diagram is different.

Wait — could it be that ∠MON = ∠MOP - ∠NOP?

Suppose M-O-P, and N is between M and P → then ∠MOP = ∠MON + ∠NOP

So again: 3x + 3 = 63 + (9x - 15)

Same problem.

Unless the labels are swapped.

Wait — maybe ∠MOP = 3x + 3, ∠NOP = 9x - 15, and ∠MON = 63°

But if N is between M and P, then ∠MOP = ∠MON + ∠NOP → same issue.

Wait — unless ∠MOP is actually ∠MON + ∠NOP → so:

$$
3x + 3 = 63 + (9x - 15) \\
3x + 3 = 9x + 48 \\
-6x = 45 \\
x = -7.5
$$

Still invalid.

But you got x = 6.25

Let’s try:
If instead, ∠MOP = ∠MON + ∠NOP → then:
3x + 3 = 63 + (9x - 15) → no.

Wait — maybe the expression is reversed.

Wait — perhaps the problem says:
- ∠MOP = 3x + 3
- ∠NOP = 9x - 15
- ∠MON = 63°

And the diagram shows that ∠MOP = ∠MON + ∠NOP → so:
3x + 3 = 63 + (9x - 15) → still bad.

Wait — unless ∠MOP = ∠MON + ∠NOP → so:

3x + 3 = 63 + (9x - 15) → no.

Wait — maybe it's ∠MOP = ∠MON + ∠NOP → but ∠MOP = 3x+3, ∠MON = 63, ∠NOP = 9x-15

So:
3x + 3 = 63 + (9x - 15) → 3x + 3 = 9x + 48 → x = -7.5 → impossible.

But you solved:
3x + 3 + 9x - 15 = 63 → 12x - 12 = 63 → x = 75/12 = 6.25

That suggests:
∠MOP + ∠NOP = ∠MON → which would mean they are on opposite sides.

But geometrically, that doesn't make sense unless it's a reflex angle.

Alternatively, perhaps the diagram has ray ON between OM and OP, so:
∠MOP = ∠MON + ∠NOP

But then ∠MOP must be greater than ∠MON.

But 3x + 3 = ? 63 + (9x - 15) → left side smaller than right for positive x.

So unless the expressions are switched.

Wait — maybe ∠MOP = 9x - 15, and ∠NOP = 3x + 3?

But the problem says:
- m∠MOP = (3x + 3)°
- m∠NOP = (9x - 15)°
- m∠MON = 63°

Perhaps the diagram shows that ∠MON = ∠MOP - ∠NOP?

Then:
63 = (3x + 3) - (9x - 15) = 3x + 3 - 9x + 15 = -6x + 18

So:
63 = -6x + 18 → 6x = 18 - 63 = -45 → x = -7.5 → again negative.

Not possible.

Wait — maybe it's ∠MOP = ∠MON + ∠NOP → so:
3x + 3 = 63 + (9x - 15) → same issue.

But you got x = 6.25, and then:
m∠MOP = 3(6.25) + 3 = 18.75 + 3 = 21.75°
m∠NOP = 9(6.25) - 15 = 56.25 - 15 = 41.25°
Sum = 21.75 + 41.25 = 63° → oh! That equals ∠MON

So you assumed:
∠MOP + ∠NOP = ∠MON → 63°

But that would only make sense if rays are arranged such that ∠MOP and ∠NOP are adjacent and together form ∠MON.

But that would require P and N to be on the same side.

Wait — maybe the diagram has ray OP and ON on the same side, and OM on the other.

But typically, ∠MON is between M and N.

Wait — perhaps the correct interpretation is that ∠MON = ∠MOP + ∠PON → but PON is same as NOP.

But you have:
∠MOP = 3x+3
∠NOP = 9x-15
Sum = 3x+3 + 9x-15 = 12x -12

Set equal to ∠MON = 63°

So:
12x - 12 = 63 → 12x = 75 → x = 6.25

Then:
- ∠MOP = 3(6.25) + 3 = 18.75 + 3 = 21.75°
- ∠NOP = 9(6.25) - 15 = 56.25 - 15 = 41.25°
- Sum = 63° → matches ∠MON

So the assumption is: ∠MON = ∠MOP + ∠NOP → meaning that P is between M and N.

So ray OP is between OM and ON.

So the diagram is: M-O-P-N

Then ∠MON = ∠MOP + ∠PON = ∠MOP + ∠NOP

So yes, valid.

So your solution is correct.

✔ x = 6.25

Then:
- m∠MOP = 21.75°
- m∠NOP = 41.25°

You wrote:
```
x = 6.25
m∠MOP = 21.75
m∠NOP = 41.25
```

✔ Correct.

---

Problem 5:

> If $ m\angle MNO = (5x + 3)^\circ $, $ m\angle ONP = (3x - 4)^\circ $, and ∠MNP is a right angle, find each measure.

So ∠MNP = 90°

And ∠MNP = ∠MNO + ∠ONP → since O is on MN and NP

So:
$$
(5x + 3) + (3x - 4) = 90 \\
8x - 1 = 90 \\
8x = 91 \\
x = 11.375
$$

You wrote:
```
5x+3 + 3x-4 = 90
8x -1 = 90
8x = 91
x = ?
```

Then:
- x = 11.375
- m∠MNO = 5(11.375) + 3 = 56.875 + 3 = 59.875°
- m∠ONP = 3(11.375) - 4 = 34.125 - 4 = 30.125°

Sum: 59.875 + 30.125 = 90° → correct.

So your work is correct.

✔ x = 11.375

---

Final Summary:

| Problem | Your Work | Correctness |
|--------|----------|-----------|
| 1a | B | ✔ |
| 1b | BE | ✘ (should be BD and BC) |
| 1c | ∠EBD | ✔ |
| 1d | 90°, 90°, 180° | ✔ |
| 1g | DBF | ✘ (should be BF) |
| 1h | 36° | ✘ (should be 144°) |
| 1i | 36° | ✘ (should be 27°) |
| 2 | x = 6 | ✔ |
| 3 | x = 4, ∠EFH = 21°, ∠EFG = 83° | ✔ |
| 4 | x = 6.25, ∠MOP = 21.75°, ∠NOP = 41.25° | ✔ |
| 5 | x = 11.375, ∠MNO = 59.875°, ∠ONP = 30.125° | ✔ |

---

Corrections Needed:

Problem 1:
- b) Sides of ∠1: BD and BC
- g) Angle bisector: BF
- h) m∠EBC = 144°
- i) m∠ABE = 27°

All others are correct.

---

✔ Final Answer:
Most of your work is correct except for parts 1b, 1g, 1h, and 1i. Please revise those.