Geometry worksheet with problems on angle relationships and algebraic equations.

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 can see that:
- $ \angle CDF $ and $ \angle FDE $ are adjacent angles forming $ \angle CDE $.
- So:
$$
m\angle CDF + m\angle FDE = m\angle CDE
$$

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

Simplify left side:
$$
8x + 12 = 10x - 18
$$

Solve for $ x $:
$$
12 + 18 = 10x - 8x \\
30 = 2x \\
x = 15
$$

Now plug back to find 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 $

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

> ⚠️ Note: The image shows $ m\angle CDF = 140^\circ $, which is incorrect. It should be 59°.

---

Problem 7:

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

From the diagram:
- Angles $ \angle LMP $, $ \angle NMP $, and $ \angle NML $ are around point $ M $
- But $ \angle NML $ is a straight angle? Wait — no, look carefully.

Actually, from the diagram:
- Ray $ ML $, $ MP $, and $ MN $ originate from point $ M $
- $ \angle NML = 137^\circ $, which is the angle between $ MN $ and $ ML $
- $ \angle LMP $ and $ \angle NMP $ are parts of this larger angle?

Wait — let’s interpret correctly.

The key is that $ \angle NML $ is composed of $ \angle NMP $ and $ \angle PML $ (i.e., $ \angle LMP $), so:

$$
m\angle NML = m\angle NMP + m\angle LMP
$$

We are told:
- $ m\angle LMP = m\angle NMP + 11 $
- $ m\angle NML = 137^\circ $

So substitute:
$$
m\angle NMP + (m\angle NMP + 11) = 137 \\
2(m\angle NMP) + 11 = 137 \\
2(m\angle NMP) = 126 \\
m\angle NMP = 63^\circ
$$

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

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

This matches what's written in the image — correct.

---

Problem 8:

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

From the diagram:
- $ \angle ABC $ is made up of $ \angle ABD $ and $ \angle DBC $
- So:
$$
m\angle ABC = m\angle ABD + m\angle DBC
$$

Let $ m\angle ABD = x $. Then:
- $ m\angle ABC = 3x - 1 $ (one degree less than three times)
- Also: $ m\angle ABC = x + 47 $

Set equal:
$$
3x - 1 = x + 47 \\
3x - x = 47 + 1 \\
2x = 48 \\
x = 24
$$

So:
- $ m\angle ABD = 24^\circ $
- $ m\angle ABC = 24 + 47 = 71^\circ $

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

> ⚠️ Image shows $ m\angle ABD = 71^\circ $, $ m\angle ABC = 71^\circ $ — this is wrong. Correct values are 24° and 71°, respectively.

---

Problem 9:

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

From the diagram:
- Point $ Q $ with rays $ QR $, $ QS $, $ QT $, $ QP $
- $ \angle PQT $ is split by $ QS $ into $ \angle PQS $ and $ \angle SQT $
- Since $ QS $ bisects $ \angle PQT $, then:
$$
m\angle PQS = m\angle SQT = \frac{1}{2} m\angle PQT
$$

Also, $ \angle RQT $ is made of $ \angle RQS $ and $ \angle SQT $

So:
$$
m\angle RQT = m\angle RQS + m\angle SQT
$$

We know:
- $ m\angle RQT = 8x - 25 $
- $ m\angle RQS = 112^\circ $
- $ m\angle SQT = \frac{1}{2}(9x + 34) $

So:
$$
8x - 25 = 112 + \frac{1}{2}(9x + 34)
$$

Multiply both sides by 2 to eliminate fraction:
$$
2(8x - 25) = 2(112) + (9x + 34) \\
16x - 50 = 224 + 9x + 34 \\
16x - 50 = 258 + 9x \\
16x - 9x = 258 + 50 \\
7x = 308 \\
x = 44
$$

Now compute angles:

- $ m\angle PQT = 9(44) + 34 = 396 + 34 = 430^\circ $? ✘ That's impossible!

Wait — 430° is not possible for an angle. Something went wrong.

Let’s double-check.

Wait — maybe I misread the diagram.

Wait: $ m\angle RQT = 8x - 25 $, and $ m\angle RQS = 112^\circ $, and $ m\angle SQT = ? $

But if $ m\angle RQT = 8x - 25 $, and $ m\angle RQS = 112^\circ $, and $ \angle RQS $ is part of $ \angle RQT $, then $ m\angle SQT = (8x - 25) - 112 $

But also, since $ QS $ bisects $ \angle PQT $, then:
$$
m\angle PQS = m\angle SQT = \frac{1}{2}(9x + 34)
$$

So:
$$
m\angle SQT = (8x - 25) - 112 = 8x - 137
$$

But also:
$$
m\angle SQT = \frac{1}{2}(9x + 34)
$$

Set equal:
$$
8x - 137 = \frac{1}{2}(9x + 34)
$$

Multiply both sides by 2:
$$
16x - 274 = 9x + 34 \\
16x - 9x = 34 + 274 \\
7x = 308 \\
x = 44
$$

Same as before.

Now check $ m\angle PQT = 9(44) + 34 = 396 + 34 = 430^\circ $ → still too big.

But wait — maybe the diagram has rays going in different directions. Perhaps $ \angle PQT $ is not the full angle?

Alternatively, maybe $ \angle RQT $ is not adjacent to $ \angle RQS $? Let’s re-express.

Wait — perhaps the labeling is such that $ R $, $ S $, $ T $ are all on the same side?

Wait — maybe $ \angle RQT $ includes $ \angle RQS $ and $ \angle SQT $, so yes, $ \angle RQT = \angle RQS + \angle SQT $

But if $ x = 44 $, then:

- $ m\angle RQT = 8(44) - 25 = 352 - 25 = 327^\circ $
- $ m\angle RQS = 112^\circ $
- So $ m\angle SQT = 327 - 112 = 215^\circ $

But $ m\angle SQT = \frac{1}{2}(9x + 34) = \frac{1}{2}(396 + 34) = \frac{430}{2} = 215^\circ $ ✔

So mathematically consistent, but angles over 180° are possible in geometry, though unusual.

But wait — if $ \angle PQT = 430^\circ $, that's more than a full circle! Impossible.

Ah — here’s the issue: $ \angle PQT $ must be less than 360°, but 430° > 360° → invalid.

So either the problem has a typo, or we misinterpreted.

Wait — maybe $ m\angle PQT = (9x + 34)^\circ $ is not the total angle?

Wait — the notation: $ m\angle PQT $ is the angle from $ P $ to $ T $ via $ Q $, so it should be the angle at $ Q $ between points $ P $, $ Q $, $ T $

If $ QS $ bisects $ \angle PQT $, then $ \angle PQS = \angle SQT $

And $ m\angle PQT = \angle PQS + \angle SQT = 2 \times \angle SQT $

We have:
- $ m\angle RQT = 8x - 25 $
- $ m\angle RQS = 112^\circ $
- So $ m\angle SQT = (8x - 25) - 112 = 8x - 137 $

But $ m\angle SQT = \frac{1}{2} m\angle PQT = \frac{1}{2}(9x + 34) $

So:
$$
8x - 137 = \frac{1}{2}(9x + 34)
$$

As before → $ x = 44 $

Now:
- $ m\angle PQT = 9(44) + 34 = 396 + 34 = 430^\circ $ → invalid

So contradiction.

Wait — unless the angle is measured differently.

Maybe $ \angle PQT $ is reflex? But usually not.

Alternatively, perhaps the expression $ m\angle PQT = (9x + 34)^\circ $ is meant to be the non-reflex angle, so less than 180°?

But even then, 430° is impossible.

Wait — perhaps there's a typo in the problem.

Alternatively, maybe $ m\angle RQT = (8x - 25)^\circ $ is actually less than $ m\angle RQS $? But $ m\angle RQS = 112^\circ $, and $ 8x - 25 $ with $ x=44 $ is 327°, so $ RQT $ is bigger — okay.

But $ \angle PQT = 430^\circ $ is impossible.

Wait — unless the diagram shows $ \angle PQT $ as the smaller angle? But the expression gives 430°.

Alternative idea: Maybe $ m\angle PQT = (9x + 34)^\circ $ is not the full angle, but only one part?

No — it says $ m\angle PQT $, so it should be the whole angle.

Wait — perhaps the variable is defined differently.

Wait — maybe $ m\angle RQT $ is not $ 8x - 25 $, but rather $ m\angle RQS = 112^\circ $, and $ m\angle SQT = ? $

But the problem says:
> $ m\angle RQT = (8x - 25)^\circ $
> $ m\angle PQT = (9x + 34)^\circ $
> $ m\angle RQS = 112^\circ $

And $ QS $ bisects $ \angle PQT $

So $ \angle PQS = \angle SQT = \frac{1}{2}(9x + 34) $

Now, $ \angle RQT = \angle RQS + \angle SQT $

So:
$$
8x - 25 = 112 + \frac{1}{2}(9x + 34)
$$

Which we solved → $ x = 44 $

But leads to $ m\angle PQT = 430^\circ $ → invalid.

So unless the problem allows reflex angles, it’s invalid.

But in standard geometry, angles are taken as the smaller one unless specified.

So likely, there is a typo in the problem.

Alternatively, maybe $ m\angle RQT = (8x - 25)^\circ $ is supposed to be $ m\angle RQS = 8x - 25 $? But no, it says $ m\angle RQS = 112^\circ $

Wait — maybe the labels are off.

Alternatively, maybe $ m\angle PQT = (9x + 34)^\circ $ is the reflex angle? Then the actual smaller angle is $ 360 - (9x + 34) $

But that complicates things.

Alternatively, perhaps $ m\angle RQT = (8x - 25)^\circ $ is not the angle from R to T, but something else.

Given the confusion, and since $ x = 44 $ gives $ m\angle PQT = 430^\circ $, which is impossible, perhaps the problem has a typo.

But let's assume the math is correct and proceed.

So:
- $ x = 44 $
- $ m\angle PQT = 9(44) + 34 = 430^\circ $
- $ m\angle PQS = m\angle SQT = 215^\circ $
- $ m\angle RQT = 8(44) - 25 = 327^\circ $
- $ m\angle RQS = 112^\circ $
- $ m\angle SQT = 327 - 112 = 215^\circ $ ✔

So although $ \angle PQT = 430^\circ $ is geometrically impossible (as it exceeds 360°), perhaps it's intended as a reflex angle.

But typically, angle measures are less than 360°.

Alternatively, maybe the expression is $ m\angle PQT = (9x + 34)^\circ $, but it's actually not the total angle.

Wait — another possibility: perhaps $ \angle PQT $ is composed of $ \angle PQR $ and $ \angle RQT $, but that’s not indicated.

Given the ambiguity, and since the algebra works out, we’ll accept $ x = 44 $, but note the inconsistency.

✔ Answer (assuming the math is correct despite large angles):
- $ x = 44 $
- $ m\angle PQT = 430^\circ $
- $ m\angle PQS = 215^\circ $
- $ m\angle SQT = 215^\circ $
- $ m\angle RQS = 112^\circ $
- $ m\angle RQT = 327^\circ $

But this seems problematic.

Wait — perhaps the diagram shows $ \angle RQT $ as a small angle, and $ \angle RQS = 112^\circ $, so $ \angle SQT = m\angle RQT - 112^\circ $, but if $ m\angle RQT < 112^\circ $, then negative — impossible.

So $ m\angle RQT > 112^\circ $

But $ 8x - 25 > 112 $ → $ 8x > 137 $ → $ x > 17.125 $

With $ x = 44 $, it's fine.

But again, $ m\angle PQT = 430^\circ $ is invalid.

So likely, the problem has a typo.

Perhaps $ m\angle PQT = (9x + 34)^\circ $ is meant to be $ m\angle PQS = (9x + 34)^\circ $? But that doesn't make sense.

Alternatively, maybe $ m\angle PQT = (9x + 34)^\circ $ is not the total, but $ m\angle SQT = (9x + 34)^\circ $? No, the problem says $ m\angle PQT $.

Given time, we'll move on.

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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 = (8x + 13)^\circ $
- $ m\angle CDF = 42^\circ $? Wait — $ \angle CDF $? But $ F $ is not labeled.

Wait — the diagram shows $ C $, $ O $, $ E $, $ D $, $ F $? Or is it $ \angle CDF $? Probably a typo.

Wait — the problem says:
> $ m\angle CDF = 42^\circ $

But $ \angle CDF $ involves $ D $, $ F $, $ C $, but $ F $ isn’t mentioned elsewhere.

Wait — looking at the diagram: probably $ F $ is a typo — should be $ \angle COF $? Or $ \angle DOF $?

Wait — the diagram shows rays $ OC $, $ OD $, $ OE $, $ OF $, etc.

Wait — the problem says:
> $ \angle COE $ is a straight angle → so $ C $, $ O $, $ E $ are colinear, with $ O $ in middle.

$ \overrightarrow{OE} $ bisects $ \angle COD $ — but $ \angle COD $ is between $ C $, $ O $, $ D $

But $ OE $ is on the opposite side of $ OC $, so how can it bisect $ \angle COD $?

Unless $ D $ is on the other side.

Wait — perhaps $ \angle COD $ is formed by $ C $, $ O $, $ D $, and $ OE $ is the bisector.

But $ \angle COE $ is a straight angle → $ C $, $ O $, $ E $ are on a line.

So $ \angle COE = 180^\circ $

But it says $ m\angle COE = (9x - 1)^\circ $, so:
$$
9x - 1 = 180 \\
9x = 181 \\
x = \frac{181}{9} \approx 20.11
$$

But then $ m\angle DOE = (8x + 13)^\circ $

But $ \angle COE = 180^\circ $, and $ OE $ bisects $ \angle COD $

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

But $ E $ is on the extension of $ CO $, so $ \angle COE = 180^\circ $

Then $ \angle COD $ must be on the other side.

Let’s suppose:
- $ C $, $ O $, $ E $ are colinear, $ \angle COE = 180^\circ $
- $ D $ is somewhere above, so $ \angle COD $ is the angle from $ C $ to $ D $
- $ OE $ bisects $ \angle COD $ → so $ \angle COE = \angle EOD $?

But $ \angle COE = 180^\circ $, so $ \angle EOD = 180^\circ $? Then $ D $ would be on the line $ OE $, which may not be.

Contradiction.

Wait — maybe $ \angle COD $ is the angle at $ O $ between $ C $ and $ D $, and $ OE $ is the bisector.

But $ OE $ is on the straight line from $ C $ through $ O $ to $ E $, so if $ OE $ bisects $ \angle COD $, then $ D $ must be symmetric across $ OE $, but $ OE $ is a straight line.

Wait — unless $ D $ is on the other side.

Suppose:
- $ \angle COD $ is split by $ OE $ into two equal parts: $ \angle COE $ and $ \angle EOD $

But $ \angle COE = 180^\circ $, so $ \angle EOD = 180^\circ $? Only if $ D $ is on the line $ OE $, but then $ \angle COD = 360^\circ $? No.

This is confusing.

Wait — perhaps the problem means:
- $ \angle COE $ is a straight angle → $ 180^\circ $
- $ OE $ bisects $ \angle COD $, so $ \angle COE = \angle EOD $

But $ \angle COE = 180^\circ $, so $ \angle EOD = 180^\circ $, so $ D $ is on the line $ OE $, but then $ \angle COD = 360^\circ $? Not possible.

Alternatively, maybe $ \angle COD $ is the angle between $ C $ and $ D $, and $ OE $ is the bisector, so:
$$
\angle COE = \angle EOD
$$

But $ \angle COE = 180^\circ $, so $ \angle EOD = 180^\circ $, so $ D $ is on the opposite ray.

Then $ \angle COD = 360^\circ $? No.

This suggests a mistake in interpretation.

Wait — perhaps $ \angle COE $ is not $ \angle COE $, but $ \angle COE $ is the straight angle, so $ C-O-E $ are colinear.

Then $ \angle COD $ is the angle from $ C $ to $ D $, and $ OE $ bisects it.

So $ \angle COE $ is part of $ \angle COD $? But $ \angle COE = 180^\circ $, so $ \angle COD \geq 180^\circ $

But $ OE $ bisects $ \angle COD $, so:
$$
\angle COE = \angle EOD = \frac{1}{2} \angle COD
$$

But $ \angle COE = 180^\circ $, so $ \angle COD = 360^\circ $, which is a full circle — possible if $ D $ is on the same line as $ C $ and $ E $, but then $ \angle COD = 180^\circ $ if $ D $ is on the opposite side.

Wait — if $ C $, $ O $, $ E $ are colinear, and $ D $ is on the opposite side, then $ \angle COD = 180^\circ $, and $ OE $ cannot bisect it unless $ D $ is on the line.

I think there's a typo.

Perhaps $ \angle COE $ is not the straight angle, but $ \angle COE $ is the straight angle, and $ OE $ bisects $ \angle COD $, so $ \angle COE = \angle EOD $, but $ \angle COE = 180^\circ $, so $ \angle EOD = 180^\circ $, so $ D $ is on the line $ OE $, but then $ \angle COD = 180^\circ $, and $ OE $ is the bisector, so $ \angle COE = \angle EOD = 90^\circ $? Contradiction.

I think the problem has a typo.

Alternatively, perhaps $ \angle COE $ is not the straight angle, but $ \angle COE $ is the straight angle, so $ 180^\circ $, and $ OE $ bisects $ \angle COD $, so $ \angle COE = \angle EOD $, but $ \angle COE = 180^\circ $, so $ \angle EOD = 180^\circ $, so $ D $ is on the line $ OE $, and $ \angle COD = 360^\circ $? No.

Given the complexity and time, and since the image shows answers, perhaps we can use them.

But without a clear diagram, it's hard.

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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 = 44 $
- $ m\angle PQT = 430^\circ $ ← likely error; should be checked
- $ m\angle PQS = 215^\circ $
- $ m\angle SQT = 215^\circ $
- $ m\angle RQS = 112^\circ $
- $ m\angle RQT = 327^\circ $

#### Problem 10:
Incomplete due to unclear diagram and possible typos.

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✔ Corrected Answers (based on logic):

| Problem | Answer |
|--------|--------|
| 6 | $ x = 15 $, $ \angle CDF = 59^\circ $, $ \angle FDE = 73^\circ $, $ \angle CDE = 132^\circ $ |
| 7 | $ \angle LMP = 74^\circ $, $ \angle NMP = 63^\circ $ |
| 8 | $ \angle ABD = 24^\circ $, $ \angle ABC = 71^\circ $ |
| 9 | $ x = 44 $, but $ \angle PQT = 430^\circ $ is invalid; likely typo in problem |
| 10 | Needs clarification |

Let me know if you want help with Problem 10 or corrections.