SOLUTION: Angle Addition Postulate Geometry Basics Worksheet ... - Free Printable

Let's solve each problem step by step, based on the given information and diagrams.

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Problem 6:

Given:
- $ m\angle CDF = (3x + 14)^\circ $
- $ m\angle FDE = (5x - 2)^\circ $
- $ m\angle CDE = (10x - 18)^\circ $

From the diagram, we see that point $ D $ is the vertex, and rays $ DC $, $ DF $, and $ DE $ form angles such that:

$$
\angle CDF + \angle FDE = \angle CDE
$$

So,

$$
(3x + 14) + (5x - 2) = 10x - 18
$$

Simplify left side:

$$
8x + 12 = 10x - 18
$$

Subtract $ 8x $ from both sides:

$$
12 = 2x - 18
$$

Add 18 to both sides:

$$
30 = 2x \Rightarrow x = 15
$$

Now plug back into each angle:

- $ m\angle CDF = 3(15) + 14 = 45 + 14 = 59^\circ $
- $ m\angle FDE = 5(15) - 2 = 75 - 2 = 73^\circ $
- $ m\angle CDE = 10(15) - 18 = 150 - 18 = 132^\circ $

✔ Check: $ 59 + 73 = 132 $ → Correct!

Answer:
- $ x = 15 $
- $ m\angle CDF = 59^\circ $
- $ m\angle FDE = 73^\circ $
- $ m\angle CDE = 132^\circ $

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Problem 7:

Given:
- $ m\angle LMP $ is 11° more than $ m\angle NMP $
- $ m\angle NML = 137^\circ $

From the diagram, rays $ ML $, $ MP $, and $ MN $ originate from point $ M $. Since $ \angle NML = 137^\circ $, and $ MP $ lies between $ ML $ and $ MN $, then:

$$
\angle LMP + \angle NMP = \angle NML = 137^\circ
$$

Let $ m\angle NMP = x $. Then $ m\angle LMP = x + 11 $

So:

$$
x + (x + 11) = 137 \\
2x + 11 = 137 \\
2x = 126 \Rightarrow x = 63
$$

Then:
- $ m\angle NMP = 63^\circ $
- $ m\angle LMP = 63 + 11 = 74^\circ $

✔ Check: $ 63 + 74 = 137 $ → Correct!

Answer:
- $ m\angle LMP = 74^\circ $
- $ m\angle NMP = 63^\circ $

---

Problem 8:

Given:
- $ m\angle ABC $ is one degree less than three times $ m\angle ABD $
- $ m\angle DBC = 47^\circ $

From the diagram, ray $ BD $ divides $ \angle ABC $, so:

$$
\angle ABC = \angle ABD + \angle DBC
$$

Let $ m\angle ABD = x $. Then:

- $ m\angle ABC = 3x - 1 $ (one degree less than three times)
- $ m\angle DBC = 47^\circ $

So:

$$
\angle ABC = \angle ABD + \angle DBC \\
3x - 1 = x + 47
$$

Subtract $ x $ from both sides:

$$
2x - 1 = 47 \\
2x = 48 \Rightarrow x = 24
$$

Then:
- $ m\angle ABD = 24^\circ $
- $ m\angle ABC = 3(24) - 1 = 72 - 1 = 71^\circ $

✔ Check: $ 24 + 47 = 71 $ → Correct!

Answer:
- $ m\angle ABD = 24^\circ $
- $ m\angle ABC = 71^\circ $

---

Problem 9:

Given:
- $ \overrightarrow{QH} $ bisects $ \angle PQT $
- $ m\angle PQH = (8x - 25)^\circ $
- $ m\angle PQT = (9x + 34)^\circ $
- $ m\angle RQS = 112^\circ $

We are to find $ x $, and then the measures of angles $ PQH $, $ PQT $, and $ RQS $. But wait — there’s a typo in your image? It says “find each measure” but only gives $ RQS = 112^\circ $. Let's analyze.

Since $ QH $ bisects $ \angle PQT $, it splits it into two equal parts:

$$
m\angle PQH = m\angle HQT
$$

And:

$$
m\angle PQT = 2 \cdot m\angle PQH
$$

So:

$$
9x + 34 = 2(8x - 25)
$$

Solve:

$$
9x + 34 = 16x - 50 \\
34 + 50 = 16x - 9x \\
84 = 7x \Rightarrow x = 12
$$

Now compute:

- $ m\angle PQH = 8(12) - 25 = 96 - 25 = 71^\circ $
- $ m\angle PQT = 9(12) + 34 = 108 + 34 = 142^\circ $
- $ m\angle RQS = 112^\circ $ (given)

Wait — but what about $ RQS $? Is it related?

Looking at the diagram (though not visible), likely points $ R $, $ Q $, $ S $, etc., form a straight line or other configuration. But since $ RQS = 112^\circ $ is given, and no equation connects it directly, perhaps it's just provided as extra info or part of another angle.

But if $ RQS = 112^\circ $, and assuming it's adjacent or vertical, but unless more info, we can't relate it directly. However, since the problem says "find each measure", and only asks for $ x $, $ PQH $, $ PQT $, and $ RQS $, and $ RQS $ is already given, maybe it's just to confirm.

But let’s assume the goal is to find the others.

So far:
- $ x = 12 $
- $ m\angle PQH = 71^\circ $
- $ m\angle PQT = 142^\circ $
- $ m\angle RQS = 112^\circ $

But is there a relationship between $ RQS $ and the rest? Possibly, if $ RQS $ and $ PQT $ are vertical angles or supplementary?

Alternatively, maybe $ RQS $ is meant to be used to find $ x $? But no equation links them unless the diagram shows that $ RQS $ and $ PQT $ are vertical or supplementary.

Wait — perhaps $ \angle RQS $ is a typo? Or maybe it's $ \angle RQS $ and $ \angle PQT $ are vertical angles?

If they are vertical angles, then $ \angle RQS = \angle PQT $, but $ 112 \ne 142 $, so not equal.

If they are supplementary? $ 112 + 142 = 254 $ ≠ 180 → No.

Alternatively, perhaps $ \angle RQS $ is adjacent to $ \angle PQT $? Without the diagram, it's hard.

But since $ \angle RQS = 112^\circ $ is given, and no relation is specified, maybe it's just an extra piece of data — or perhaps the problem has a typo.

Alternatively, maybe $ \angle RQS $ is meant to be $ \angle RQP $ or something else.

But based on the information:

We solved:
- $ x = 12 $
- $ m\angle PQH = 71^\circ $
- $ m\angle PQT = 142^\circ $
- $ m\angle RQS = 112^\circ $ (given)

So unless further relations exist, these are the answers.

But perhaps the diagram shows that $ \angle RQS $ and $ \angle PQT $ are adjacent forming a straight line? But 112 + 142 = 254 > 180 → No.

Alternatively, maybe $ \angle RQS $ is actually $ \angle RQP $? Not sure.

But since $ \angle RQS $ is given as 112°, and no variable depends on it, we’ll assume it’s just listed.

Answer:
- $ x = 12 $
- $ m\angle PQH = 71^\circ $
- $ m\angle PQT = 142^\circ $
- $ m\angle RQS = 112^\circ $

---

Problem 10:

Given:
- $ \angle COE $ is a straight angle → $ 180^\circ $
- $ \overrightarrow{OE} $ bisects $ \angle COD $
- $ m\angle COE = (9x - 1)^\circ $
- $ m\angle DOE = (6x + 13)^\circ $
- $ m\angle CDF = 42^\circ $ ← Wait, this seems off.

Wait — $ \angle CDF $? But point $ F $ isn't mentioned in this diagram. Probably a typo.

Likely, it should be $ m\angle CDF $ is a typo. Looking at the diagram (implied), probably it's $ m\angle CDF $ → maybe $ m\angle CDO $ or $ m\angle CDF $ is not relevant.

Wait — re-read:

> If $ \angle COE $ is a straight angle, $ OE $ bisects $ \angle COD $, $ m\angle COE = (9x - 1)^\circ $, $ m\angle DOE = (6x + 13)^\circ $, and $ m\angle CDF = 42^\circ $

But $ \angle CDF $ involves point $ F $, which is not in $ \angle COE $. Likely a typo.

Possibly $ m\angle CDF $ should be $ m\angle CDO $? Or $ m\angle CEF $? Or perhaps $ m\angle CDF $ is meant to be $ m\angle CDE $? But without diagram, hard.

Wait — perhaps $ \angle COE $ is straight → $ 180^\circ $

So:

$$
m\angle COE = 180^\circ = (9x - 1)^\circ
$$

So:

$$
9x - 1 = 180 \Rightarrow 9x = 181 \Rightarrow x = \frac{181}{9} \approx 20.11
$$

But that’s messy.

Alternatively, maybe $ m\angle COE = 180^\circ $, so:

$$
9x - 1 = 180 \Rightarrow x = \frac{181}{9} \approx 20.11
$$

But also given $ m\angle DOE = (6x + 13)^\circ $

And $ OE $ bisects $ \angle COD $, so:

$$
\angle COE = \angle COD + \angle DOE?
$$

Wait — no. Let's clarify.

Point $ O $ is vertex. Rays: $ OC $, $ OD $, $ OE $

Given: $ \angle COE $ is a straight angle → $ 180^\circ $

Also, $ OE $ bisects $ \angle COD $ → So $ \angle COE $ includes $ \angle COD $? That doesn’t make sense unless $ E $ is on the extension.

Wait — better interpretation:

Assume $ \angle COE $ is a straight angle → points $ C $, $ O $, $ E $ are collinear, with $ O $ between $ C $ and $ E $, so $ \angle COE = 180^\circ $

Now, $ OE $ bisects $ \angle COD $ → This means $ \angle COD $ is split by $ OE $ into two equal parts: $ \angle COE $ and $ \angle EOD $? But $ \angle COE $ is already 180°, so $ \angle COD $ would have to be larger than 180° → impossible.

Contradiction.

Alternative: Maybe $ OE $ bisects $ \angle COD $, meaning $ \angle COE = \angle EOD $, but $ \angle COE $ is straight → 180°, so $ \angle COE = 180^\circ $, $ \angle EOD = 180^\circ $? Impossible.

Wait — likely mislabeling.

Perhaps $ \angle COE $ is not the angle being bisected.

Wait — the statement says:

> If $ \angle COE $ is a straight angle, $ OE $ bisects $ \angle COD $, $ m\angle COE = (9x - 1)^\circ $, $ m\angle DOE = (6x + 13)^\circ $, and $ m\angle CDF = 42^\circ $

Ah! Here’s the issue: $ m\angle COE = (9x - 1)^\circ $, but $ \angle COE $ is a straight angle → so:

$$
9x - 1 = 180 \Rightarrow 9x = 181 \Rightarrow x = \frac{181}{9} \approx 20.11
$$

But then $ m\angle DOE = 6x + 13 = 6(181/9) + 13 = (1086/9) + 13 = 120.666... + 13 = 133.666^\circ $

But if $ OE $ bisects $ \angle COD $, then $ \angle COE = \angle EOD $? But $ \angle COE = 180^\circ $, so $ \angle EOD = 180^\circ $? No.

Wait — confusion in labeling.

Let’s suppose:

- $ \angle COE $ is a straight angle → $ 180^\circ $
- So $ m\angle COE = 180^\circ $
- Given: $ m\angle COE = (9x - 1)^\circ $ → So $ 9x - 1 = 180 \Rightarrow x = 181/9 \approx 20.11 $

But then $ m\angle DOE = 6x + 13 = 6(181/9) + 13 = (1086)/9 + 13 = 120.666 + 13 = 133.666^\circ $

Now, $ OE $ bisects $ \angle COD $ → So $ \angle COE = \angle EOD $? But $ \angle COE = 180^\circ $, so $ \angle EOD = 180^\circ $? But $ \angle DOE = 133.666^\circ $, contradiction.

So something is wrong.

Wait — perhaps $ \angle COE $ is not the full straight angle? But the problem says "if $ \angle COE $ is a straight angle" → so it must be 180°.

Another possibility: The expression $ m\angle COE = (9x - 1)^\circ $ is incorrect? But it's written.

Alternatively, maybe $ \angle COE $ is not the straight angle? But the problem says it is.

Wait — perhaps the label is wrong.

Let me try a different interpretation.

Suppose:

- $ \angle COE $ is a straight angle → $ 180^\circ $
- So $ m\angle COE = 180^\circ $
- But given: $ m\angle COE = (9x - 1)^\circ $ → so $ 9x - 1 = 180 \Rightarrow x = 181/9 \approx 20.11 $

Then $ m\angle DOE = 6x + 13 = 6*(181/9) + 13 = (1086/9) + 13 = 120.666 + 13 = 133.666^\circ $

Now, $ OE $ bisects $ \angle COD $ → So $ \angle COE = \angle EOD $? But $ \angle COE = 180^\circ $, $ \angle DOE = 133.666^\circ $ → not equal.

Contradiction.

Unless $ \angle COD $ is not the same as $ \angle COE $.

Wait — perhaps $ \angle COD $ is the angle between $ C $, $ O $, $ D $, and $ OE $ is inside it.

But $ \angle COE $ is straight → so $ C $, $ O $, $ E $ are collinear.

So $ E $ is on the opposite side of $ C $ through $ O $.

Then $ D $ is somewhere.

$ OE $ bisects $ \angle COD $ → So $ \angle COE = \angle EOD $? But $ \angle COE = 180^\circ $, so $ \angle EOD = 180^\circ $? But $ D $ is a point, so $ \angle EOD $ is part of it.

Wait — unless $ D $ is on the line $ CE $, but then $ \angle COD $ is 180°, and $ OE $ bisects it → $ \angle COE = \angle EOD = 90^\circ $? But $ \angle COE = 180^\circ $, so can't be.

This suggests inconsistency.

Wait — perhaps $ \angle COE $ is not the straight angle, but rather $ \angle COE $ is part of the straight angle.

Wait — reread:

> If $ \angle COE $ is a straight angle

So $ \angle COE = 180^\circ $

But then $ m\angle COE = (9x - 1)^\circ $ → so $ 9x - 1 = 180 \Rightarrow x = 181/9 $

But then $ m\angle DOE = 6x + 13 = 6*(181/9) + 13 = 1086/9 + 13 = 120.666 + 13 = 133.666^\circ $

Now, $ OE $ bisects $ \angle COD $ → so $ \angle COE = \angle EOD $? But $ \angle COE = 180^\circ $, $ \angle DOE = 133.666^\circ $ → not equal.

So only way this works is if $ \angle COE $ is not the straight angle.

Wait — maybe the problem says $ \angle COE $ is a straight angle, but the expression $ m\angle COE = (9x - 1)^\circ $ is for a different angle?

No — it says: $ m\angle COE = (9x - 1)^\circ $, and $ \angle COE $ is a straight angle → so $ 9x - 1 = 180 $

But then $ m\angle DOE = 6x + 13 $

And $ OE $ bisects $ \angle COD $ → so $ \angle COE = \angle EOD $? But $ \angle COE = 180^\circ $, so $ \angle EOD = 180^\circ $? But $ \angle DOE = 6x+13 = 133.666^\circ $, not 180.

So contradiction.

Unless $ \angle COD $ is not $ \angle COE $, but $ \angle COD $ is between $ C $, $ O $, $ D $, and $ OE $ is the bisector.

But if $ \angle COE = 180^\circ $, then $ E $ is on the opposite side of $ C $ through $ O $, so $ \angle COE $ is straight.

Then $ D $ is somewhere in the plane.

Let $ \angle COD $ be the angle from $ C $ to $ D $, and $ OE $ bisects it.

But $ OE $ is along the straight line from $ O $ to $ E $, which is opposite to $ C $.

So $ OE $ is in the direction of $ E $, which is opposite to $ C $.

So if $ OE $ bisects $ \angle COD $, then $ \angle COE = \angle EOD $

But $ \angle COE $ is the angle from $ C $ to $ E $, which is 180°, so $ \angle EOD = 180^\circ $? Only if $ D $ is on the line $ OE $, but then $ \angle COD = 180^\circ $, and $ OE $ bisects it → $ \angle COE = \angle EOD = 90^\circ $? No.

Wait — if $ \angle COD $ is 180°, and $ OE $ bisects it, then $ \angle COE = \angle EOD = 90^\circ $

But the problem says $ \angle COE $ is a straight angle → 180°, so conflict.

So the only resolution is that $ \angle COE $ is not the angle being bisected.

Wait — perhaps the problem is:

> If $ \angle COE $ is a straight angle, $ OE $ bisects $ \angle COD $, $ m\angle COD = (9x - 1)^\circ $, $ m\angle DOE = (6x + 13)^\circ $, and $ m\angle CDF = 42^\circ $

But in your image, it says $ m\angle COE = (9x - 1)^\circ $, which is problematic.

Perhaps it's a typo, and it should be $ m\angle COD = (9x - 1)^\circ $

Let’s assume that.

So corrected version:

- $ \angle COE $ is a straight angle → $ 180^\circ $
- $ OE $ bisects $ \angle COD $
- $ m\angle COD = (9x - 1)^\circ $
- $ m\angle DOE = (6x + 13)^\circ $
- $ m\angle CDF = 42^\circ $ ← still problematic

But if $ OE $ bisects $ \angle COD $, then $ \angle COE = \angle EOD $? No.

Wait — $ OE $ bisects $ \angle COD $ → so $ \angle COE = \angle EOD $? But $ \angle COE $ is 180°, so $ \angle EOD = 180^\circ $? No.

Wait — perhaps $ \angle COD $ is the angle between $ C $, $ O $, $ D $, and $ OE $ is the bisector, so $ \angle COE = \angle EOD $

But $ \angle COE $ is 180°, so $ \angle EOD = 180^\circ $? Only if $ D $ is on the line $ OE $, but then $ \angle COD = 180^\circ $, and $ OE $ bisects it → $ \angle COE = \angle EOD = 90^\circ $? Contradiction.

I think there's a typo in the problem.

Perhaps it's $ \angle COD $ is a straight angle, not $ \angle COE $.

Or perhaps $ \angle COE $ is not the straight angle.

Given the confusion, and since you wrote "m∠CDF = 42°", but in this diagram, $ F $ is not present, likely it's a typo.

Perhaps it's $ m\angle CDO = 42^\circ $ or $ m\angle DOE = 42^\circ $?

But in your image, it says "m∠CDF = 42°", which is likely a mistake.

Given time, and since problems 6–9 are solvable, and 10 has likely typos, I'll stop here.

But let’s try to assume:

- $ \angle COE $ is a straight angle → 180°
- $ OE $ bisects $ \angle COD $
- $ m\angle DOE = (6x + 13)^\circ $
- $ m\angle COD = 2 \cdot m\angle DOE $ because $ OE $ bisects it → so $ m\angle COD = 2(6x + 13) = 12x + 26 $
- But $ m\angle COD $ is part of the straight angle? Not necessarily.

But if $ \angle COE $ is straight, and $ D $ is on the same side, then $ \angle COD + \angle DOE = \angle COE = 180^\circ $? Only if $ D $ is between $ C $ and $ E $.

But $ OE $ is the bisector of $ \angle COD $, so $ \angle COE = \angle EOD $

But $ \angle COE $ is 180°, so $ \angle EOD = 180^\circ $? Impossible.

So the only logical conclusion is that $ \angle COE $ is not the straight angle, but rather $ \angle COD $ is straight.

But the problem says $ \angle COE $ is straight.

Given the complexity and possible typos, and since you've already filled in some answers, I’ll summarize the correct ones.

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✔ Final Answers:

#### Problem 6:
- $ x = 15 $
- $ m\angle CDF = 59^\circ $
- $ m\angle FDE = 73^\circ $
- $ m\angle CDE = 132^\circ $

#### Problem 7:
- $ m\angle LMP = 74^\circ $
- $ m\angle NMP = 63^\circ $

#### Problem 8:
- $ m\angle ABD = 24^\circ $
- $ m\angle ABC = 71^\circ $

#### Problem 9:
- $ x = 12 $
- $ m\angle PQH = 71^\circ $
- $ m\angle PQT = 142^\circ $
- $ m\angle RQS = 112^\circ $ (given)

#### Problem 10: (Likely contains typos)
Given the inconsistencies, it's best to double-check the problem statement or diagram. But if we assume:
- $ \angle COE = 180^\circ $ → $ 9x - 1 = 180 $ → $ x = 181/9 \approx 20.11 $ → not integer.
- But likely intended: $ m\angle COD = (9x - 1)^\circ $, and $ OE $ bisects $ \angle COD $, so $ m\angle DOE = \frac{1}{2} m\angle COD $

Then:
- $ m\angle DOE = 6x + 13 $
- $ m\angle COD = 2(6x + 13) = 12x + 26 $
- Set equal to $ 9x - 1 $: $ 12x + 26 = 9x - 1 \Rightarrow 3x = -27 \Rightarrow x = -9 $ → invalid.

So no solution.

Thus, likely a typo in the problem.

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If you can provide a clearer image or correct the text for Problem 10, I can help further. Otherwise, the first four problems are solved correctly.