SOLUTION: Unit 1 Angle Addition Postulate Geometry Basics ... - Free Printable

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

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Problem 1: Use the diagram below to complete each part

#### Diagram Notes:
- Point B is the vertex where multiple rays meet.
- Rays: BA, BE, BD, BC, BF
- Given: $ \overrightarrow{BF} \perp \overrightarrow{AC} $ → So angle between BF and AC is 90°.
- Angles labeled: ∠1 = 108°, ∠5 = 36°, etc.

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#### a) Name the vertex of ∠4.

✔ Answer: B

✔️ Correct. The vertex is the common point of the two rays forming the angle. For ∠4 (which is ∠ABE), the vertex is B.

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#### b) Name the sides of ∠1.

✔ Answer: BE, BC

✔️ Correct. ∠1 is ∠EBC, so its sides are ray BE and ray BC.

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#### c) Write another name for ∠5.

✔ Answer: ∠EBD

✔️ Correct. ∠5 is formed by rays EB and DB, so it’s ∠EBD.

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#### d) Classify each angle:

- ∠FBC: ______
- ∠EBF: ______
- ∠ABC: ______

##### Let's analyze:

- ∠FBC: From the diagram, since $ \overrightarrow{BF} \perp \overrightarrow{AC} $, and AC includes points A–B–C, then ∠FBC is the angle between BF and BC.
- Since BF is perpendicular to AC, and BC is part of AC, this makes ∠FBC = 90° → Right angle
- ✔ Student wrote: 90° → ✔️ Correct classification: Right angle

- ∠EBF: This is the angle between EB and BF.
- We’re told ∠EBD = 36°, and ∠DBC = 108°, so ∠EBF might be adjacent.
- But from diagram: ∠EBF appears to be the angle from E to B to F.
- Since ∠EBD = 36°, and ∠DBF is part of the 90° right angle at B (since BF ⊥ AC), we can deduce:
- If ∠EBD = 36°, and ∠DBF = 90° − 36° = 54°, then ∠EBF = ∠EBD + ∠DBF = 36° + 54° = 90°
- So ∠EBF = 90° → Right angle
- ✔ Student wrote: 90° → ✔️ Correct

- ∠ABC: This is the angle from A to B to C, which spans the full straight line (since A–B–C are collinear).
- So ∠ABC = 180° → Straight angle
- ✔ Student wrote: 180° → ✔️ Correct

✔️ All classifications are correct.

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#### g) Name an angle bisector.

✔ Answer: DBF

Wait — DBF is not a ray; it's a combination of two rays. An angle bisector is a ray that splits an angle into two equal parts.

Looking at the diagram:
- ∠EBC is split into ∠EBD and ∠DBC.
- We’re told: ∠EBD = 36°, ∠DBC = 108° → Not equal.
- But wait! In part h), they use these values.

But if BD were the bisector, then ∠EBD = ∠DBC, but 36 ≠ 108 → So BD is not a bisector.

Wait — perhaps student meant BD? But they wrote DBF, which is confusing.

Let’s check: Is there any angle being bisected?

We know:
- ∠EBF = 90°, and ∠EBD = 36°, so ∠DBF = 54° → not equal → not bisected.
- ∠ABC = 180°, and if BF is perpendicular to AC, then ∠ABF = 90°, ∠FBC = 90° → so BF bisects ∠ABC into two 90° angles.

So BF is the angle bisector of ∠ABC.

✔ So correct answer should be: $\overrightarrow{BF}$ or just BF

✘ Student wrote: DBF → Incorrect.

💡 Correction: The angle bisector is BF, because it splits ∠ABC (180°) into two 90° angles.

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

✔ Answer: 36°

Wait — that’s wrong!

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

✘ Student wrote: 36° → ✘ Incorrect

💡 Correct Answer: 144°

Explanation: ∠EBC is the whole angle from E to B to C, which includes both ∠EBD and ∠DBC.

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

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

Given: $ m\angle EBF = 117^\circ $

From earlier, we know $ \overrightarrow{BF} \perp \overrightarrow{AC} $, so ∠ABF = 90° (since AB is on AC)

Now, ∠EBF = ∠ABE + ∠ABF

Wait — let's see the configuration:

- Points: A–B–C in a straight line
- Ray BE is going up-left
- Ray BF is going down-right (perpendicular to AC)
- So ∠EBF goes from BE to BF, passing through the center.

But ∠ABF = 90° (because BF ⊥ AC), and ∠EBF = 117°

Then ∠ABE = ∠EBF − ∠ABF = 117° − 90° = 27°

Alternatively, think of angles around point B:

- ∠ABE is the angle from A to B to E
- ∠ABF = 90° (from AB to BF)
- ∠EBF = 117° (from E to B to F)
- So the angle from A to B to E must be: ∠EBF − ∠ABF = 117° − 90° = 27°

✔ So $ m\angle ABE = 27^\circ $

✘ Student wrote: 36° → ✘ Incorrect

💡 Correct Answer: 27°

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✔ Summary of Corrections for Problem 1:

| Part | Student Answer | Correct Answer | Reason |
|------|----------------|----------------|--------|
| g) Angle bisector | DBF | BF | BF bisects ∠ABC into two 90° angles |
| h) $ m\angle EBC $ | 36° | 144° | Sum of 36° + 108° |
| i) $ m\angle ABE $ | 36° | 27° | 117° − 90° = 27° |

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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

Diagram shows three rays from K: JM, MK, KL

So:
- ∠JKL = ∠JKM + ∠MKL
- That is: $ 127^\circ = (9x - 10)^\circ + 83^\circ $

Solve:
$$
9x - 10 + 83 = 127 \\
9x + 73 = 127 \\
9x = 54 \\
x = 6
$$

✔ Student wrote:
> 9x - 10 = 44
> 9x = 54
> x = 6

Wait — where did 44 come from?
They said: 9x - 10 = 44 → but 127 - 83 = 44 → yes, that’s correct.

So:
- $ m\angle JKM = 127^\circ - 83^\circ = 44^\circ $
- Then $ 9x - 10 = 44 $
- $ 9x = 54 $
- $ x = 6 $

✔️ So x = 6 is correct.

✔ 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: E–F–G with ray FH in between.

So:
- ∠EFG = ∠EFH + ∠HFG
- $ 18x + 11 = (5x + 1) + 62 $

Solve:
$$
18x + 11 = 5x + 1 + 62 \\
18x + 11 = 5x + 63 \\
18x - 5x = 63 - 11 \\
13x = 52 \\
x = 4
$$

Now plug back:

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

Check: 21° + 62° = 83° → ✔️ Correct

Student wrote:
> 5x + 1 + 62 = 18x + 11
> 63 = 13x + 11
> 52 = 13x
> x = 4

Yes — same steps.

Then:
- $ m\angle EFH = 5(4)+1 = 21^\circ $
- $ m\angle EFG = 18(4)+11 = 83^\circ $

✔ All correct.

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Problem 4: If $ m\angle MOP = (5x - 3)^\circ $, $ m\angle MON = (9x - 15)^\circ $, and $ m\angle NOP = 60^\circ $, find each measure

Diagram: O is vertex, rays OM, ON, OP

Assuming ON is between OM and OP, then:
- ∠MOP = ∠MON + ∠NOP
- So: $ 5x - 3 = (9x - 15) + 60 $

Solve:
$$
5x - 3 = 9x - 15 + 60 \\
5x - 3 = 9x + 45 \\
-3 - 45 = 9x - 5x \\
-48 = 4x \\
x = -12
$$

Wait — negative angle? That can't be.

Let’s recheck.

Wait — maybe the order is different.

Possibility: ∠MON is larger than ∠MOP?

But ∠MOP = 5x - 3, ∠MON = 9x - 15

If ON is between OM and OP, then:
- ∠MOP = ∠MON + ∠NOP → only possible if ∠MOP > ∠MON

But 5x - 3 vs 9x - 15 → unless x is small, 9x - 15 > 5x - 3

Try solving:
$$
5x - 3 = (9x - 15) + 60 \\
5x - 3 = 9x + 45 \\
-48 = 4x \\
x = -12
$$

Negative → invalid.

Alternative: Maybe ∠MOP = ∠MON + ∠NOP, but ∠MOP is smaller? Impossible.

Wait — perhaps the labeling is different.

Maybe ∠MOP = ∠MON + ∠NOP, but the expressions suggest otherwise.

Wait — look at student’s work:

> 9x - 15 + 60 = 5x - 3
> 9x + 45 = 5x - 3
> 4x = -48
> x = -12

Same as above.

But negative x → impossible for angles.

So likely, the equation is set up backwards.

Wait — perhaps ∠MON = ∠MOP + ∠PON?

That would make more sense.

Suppose ray OP is between OM and ON.

Then:
- ∠MON = ∠MOP + ∠PON
- $ 9x - 15 = (5x - 3) + 60 $

Solve:
$$
9x - 15 = 5x + 57 \\
9x - 5x = 57 + 15 \\
4x = 72 \\
x = 18
$$

Now check:
- $ m\angle MOP = 5(18) - 3 = 90 - 3 = 87^\circ $
- $ m\angle NOP = 60^\circ $
- $ m\angle MON = 9(18) - 15 = 162 - 15 = 147^\circ $
- Check: 87 + 60 = 147 → ✔️ Correct

So correct setup is: ∠MON = ∠MOP + ∠NOP

Therefore, student incorrectly assumed ∠MOP = ∠MON + ∠NOP

✘ Student made error in setting up the equation

Correct solution:
- $ 9x - 15 = 5x - 3 + 60 $
- $ 9x - 15 = 5x + 57 $
- $ 4x = 72 $
- $ x = 18 $

Then:
- $ m\angle MOP = 87^\circ $
- $ m\angle MON = 147^\circ $

✔ Correct values.

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Problem 5: If $ m\angle MNO = (5x + 3)^\circ $, $ m\angle LNO = (3x - 4)^\circ $, and $ \angle MNL $ is a right angle, find each measure

Diagram: Point N, rays NM, NL, NO

Assume: NL and NO are rays from N, and ∠MNL = 90°

So ∠MNL = ∠MNO + ∠LNO? Or depending on position.

But ∠MNL is given as a right angle → 90°

And ∠MNO and ∠LNO share ray NO.

Possibility: Ray NO is between NM and NL → then:
- ∠MNL = ∠MNO + ∠LNO
- So: $ 5x + 3 + 3x - 4 = 90 $

Solve:
$$
8x - 1 = 90 \\
8x = 91 \\
x = 91/8 = 11.375
$$

Not nice number.

But student wrote:
> 5x + 3 + 3x - 4 = 90
> 8x - 1 = 90
> 8x = 91
> x = 91/8

Then:
- $ m\angle MNO = 5(91/8) + 3 = 455/8 + 24/8 = 479/8 = 59.875^\circ $
- $ m\angle LNO = 3(91/8) - 4 = 273/8 - 32/8 = 241/8 = 30.125^\circ $

Sum: 59.875 + 30.125 = 90 → ✔️ Correct

But usually problems have integer answers. Maybe typo?

Wait — perhaps the problem says ∠MNL is a right angle, but maybe the configuration is different.

Alternatively, could ∠MNO and ∠LNO be adjacent and add to 90°?

Yes — that’s what the student did.

So even though fractional, mathematically correct.

But let's double-check the setup.

If ∠MNL = 90°, and ray NO is inside ∠MNL, then yes:
- ∠MNL = ∠MNO + ∠LNO

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

Then:
- $ m\angle MNO = 5(91/8) + 3 = 455/8 + 24/8 = 479/8 = 59.875^\circ $
- $ m\angle LNO = 3(91/8) - 4 = 273/8 - 32/8 = 241/8 = 30.125^\circ $

✔ Correct, even if decimal.

But student wrote:
> x = 91/8
> m∠MNO = 479/8
> m∠LNO = 241/8

Which is correct.

So ✔️ Correct

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✔ Final Summary of Corrections

| Problem | Part | Student Answer | Correct Answer | Status |
|--------|------|----------------|----------------|--------|
| 1a | Vertex of ∠4 | B | B | ✔️ Correct |
| 1b | Sides of ∠1 | BE, BC | ✔️ Correct |
| 1c | Another name for ∠5 | ∠EBD | ✔️ Correct |
| 1d | ∠FBC | 90° → Right | ✔️ Correct |
| 1d | ∠EBF | 90° → Right | ✔️ Correct |
| 1d | ∠ABC | 180° → Straight | ✔️ Correct |
| 1g | Angle bisector | DBF | BF | ✘ Incorrect |
| 1h | m∠EBC | 36° | 144° | ✘ Incorrect |
| 1i | m∠ABE | 36° | 27° | ✘ Incorrect |
| 2 | x | 6 | 6 | ✔️ Correct |
| 3 | x, measures | x=4, 21°, 83° | ✔️ Correct |
| 4 | x, measures | x=-12, etc. | x=18, ∠MOP=87°, ∠MON=147° | ✘ Incorrect setup |
| 5 | x, measures | x=91/8, etc. | ✔️ Correct (even if fractional) |

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✔ Final Answers (Corrected):

#### Problem 1:
- g) BF is the angle bisector (of ∠ABC)
- h) $ m\angle EBC = 144^\circ $
- i) $ m\angle ABE = 27^\circ $

#### Problem 4:
- $ x = 18 $
- $ m\angle MOP = 87^\circ $
- $ m\angle MON = 147^\circ $

All other parts are correct.

Let me know if you'd like a clean version of the corrected homework!