Angle Addition Postulate practice worksheet with eight geometry problems involving angle measurements and algebraic expressions.

Let's solve each problem step by step using the Angle Addition Postulate, which states:

> If point $ D $ lies in the interior of $ \angle ABC $, then
> $$
> m\angle ABD + m\angle DBC = m\angle ABC
> $$

This means that when an angle is split into smaller adjacent angles, the sum of those smaller angles equals the measure of the larger angle.

---

Problem 1:

Given:
$ m\angle RUS = 32^\circ $, $ m\angle SUT = 31^\circ $
Find $ m\angle TUR $

From the diagram:
- $ \angle RUS $ and $ \angle SUT $ are adjacent angles forming $ \angle RUT $
- So, $ \angle RUT = \angle RUS + \angle SUT $

$$
m\angle TUR = 32^\circ + 31^\circ = \boxed{63^\circ}
$$

✔ Answer: $ \boxed{63^\circ} $

---

Problem 2:

Given:
$ m\angle GJI = 103^\circ $
$ m\angle GJH = 2x + 6 $, $ m\angle IJH = 3x + 7 $
Find $ x $

From the diagram:
- $ \angle GJH $ and $ \angle IJH $ together make $ \angle GJI $
- So,
$$
(2x + 6) + (3x + 7) = 103
$$
$$
5x + 13 = 103
$$
$$
5x = 90 \Rightarrow x = \boxed{18}
$$

✔ Answer: $ \boxed{x = 18} $

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Problem 3:

Given:
$ m\angle ABC = 87^\circ $
$ m\angle CBD = 5x - 9 $, $ m\angle ABD = 3x $? Wait — let's read carefully.

Wait — from the diagram:
- Point D lies between B and C?
- Actually, from the diagram:
- $ \angle ABC $ is made up of $ \angle ABD $ and $ \angle DBC $
- But labels: $ \angle CBD = 5x - 9 $, and $ \angle CBD $ is same as $ \angle DBC $
- Also $ \angle CBD $ has a label $ 5x - 9 $, and $ \angle DBC $ is shown as $ 3x $? No — wait.

Wait, the diagram shows:
- From point B, rays go to C, D, and A.
- $ \angle CBD $ is labeled $ 5x - 9 $
- $ \angle DBC $ is not labeled, but the segment from B to D splits $ \angle ABC $
- But actually, it says: $ \angle CBD = 5x - 9 $, and $ \angle CBD $ is part of $ \angle ABC $
- But also, $ \angle CBD $ seems to be one angle, and there’s another angle from B to C to D?

Wait — looking again:
- The angle at B is split into two parts:
- $ \angle CBD $ = $ 5x - 9 $
- $ \angle DBC $ = $ 3x $? No — the diagram shows:
- From B to C: ray BC
- From B to D: ray BD
- From B to A: ray BA
- So $ \angle ABC $ is composed of $ \angle ABD $ and $ \angle DBC $
- But the labels:
- $ \angle CBD $ = $ 5x - 9 $ → this is the same as $ \angle DBC $
- $ \angle CBD $ is the angle between C and D → so $ \angle CBD = 5x - 9 $
- And $ \angle DBC $ is the same thing? That can’t be.

Wait — the diagram shows:
- $ \angle CBD $ is labeled $ 5x - 9 $
- The other small angle (from C to D) is labeled $ 3x $? No — it says "3x" on the side from C to D?

Wait — no. Looking at the diagram:

- Ray BC goes down-left
- Ray BD goes down-right
- Ray BA goes right
- Angle between BC and BD is labeled $ 3x $
- Angle between BD and BA is labeled $ 5x - 9 $
- So total $ \angle ABC = \angle CBD + \angle DBA = 3x + (5x - 9) = 8x - 9 $

But we're told $ m\angle ABC = 87^\circ $

So:
$$
8x - 9 = 87 \\
8x = 96 \Rightarrow x = 12
$$

Now find $ m\angle CBD $. Since $ \angle CBD = 3x $, then:
$$
3(12) = \boxed{36^\circ}
$$

✔ Answer: $ \boxed{36^\circ} $

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Problem 4:

Given:
$ m\angle SQR = 52^\circ $
$ m\angle PQS = 3x $, $ m\angle RQP = 8x - 8 $
Find $ x $

Note: $ \angle SQR $ is the whole angle at Q, formed by points S-Q-R
Ray QP lies inside $ \angle SQR $, splitting it into:
- $ \angle PQS $ and $ \angle RQP $

So:
$$
m\angle PQS + m\angle RQP = m\angle SQR \\
3x + (8x - 8) = 52 \\
11x - 8 = 52 \\
11x = 60 \Rightarrow x = \frac{60}{11} \approx 5.45
$$

Wait — but that's not nice. Let's double-check.

Is $ \angle RQP $ the same as $ \angle QPR $? No.

Wait — the notation:
- $ \angle PQS $: from P to Q to S → that’s the left part
- $ \angle RQP $: from R to Q to P → that’s the right part
- Together: $ \angle SQR $ = $ \angle SQP + \angle PQR $, but here it's written as $ \angle PQS $ and $ \angle RQP $

Yes — $ \angle PQS $ and $ \angle RQP $ share ray QP, and together they form $ \angle SQR $

So:
$$
3x + (8x - 8) = 52 \\
11x - 8 = 52 \\
11x = 60 \Rightarrow x = \frac{60}{11}
$$

That’s unusual. Maybe typo? Or perhaps I misread.

Wait — maybe $ \angle RQP $ is the same as $ \angle PQR $? Yes, same angle.

But $ \angle SQR $ = $ \angle SQP + \angle PQR $

But $ \angle SQP = \angle PQS = 3x $, $ \angle PQR = \angle RQP = 8x - 8 $

So yes:
$$
3x + (8x - 8) = 52 \Rightarrow 11x = 60 \Rightarrow x = \frac{60}{11}
$$

But this is messy. Let’s check if units make sense.

Alternatively, could it be that $ \angle SQR $ is split into $ \angle PQS $ and $ \angle RQP $, but $ \angle RQP $ includes $ \angle PQS $? No.

Wait — the diagram shows:
- Points: P–Q–R, with S somewhere else
- Rays: from Q to P, Q to S, Q to R
- So $ \angle SQR $ is between S and R
- Ray QP is inside it → so $ \angle SQR = \angle SQP + \angle PQR $
- $ \angle SQP = \angle PQS = 3x $
- $ \angle PQR = \angle RQP = 8x - 8 $

So yes, correct.

So:
$$
3x + (8x - 8) = 52 \Rightarrow 11x = 60 \Rightarrow x = \frac{60}{11}
$$

But that seems odd for a geometry problem. Maybe typo? Or maybe I misread the expression.

Wait — could $ m\angle RQP = 8x - 8 $, and $ m\angle PQS = 3x $, and they add to 52?

Yes.

So unless the problem meant something else, we proceed.

But let’s assume it’s correct.

So:
$$
x = \frac{60}{11} \approx 5.45
$$

But let’s see if it makes sense:
- $ 3x = 3(60/11) = 180/11 ≈ 16.36^\circ $
- $ 8x - 8 = 480/11 - 8 = 480/11 - 88/11 = 392/11 ≈ 35.64^\circ $
- Sum: 16.36 + 35.64 = 52 → correct.

So mathematically correct.

But likely intended to be integer. Maybe typo in numbers?

Alternatively, maybe $ m\angle RQP = 8x - 8 $, but perhaps it should be $ 8x + 8 $? Let's not assume.

We’ll go with what's given.

✔ Answer: $ \boxed{x = \dfrac{60}{11}} $

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Problem 5:

Given:
$ m\angle DEG = 46^\circ $, $ m\angle DEF = 24x + 10 $, $ m\angle GEF = 10x - 8 $
Find $ x $

From the diagram:
- Ray EG lies between ED and EF
- So $ \angle DEF = \angle DEG + \angle GEF $

So:
$$
\angle DEG + \angle GEF = \angle DEF \\
46 + (10x - 8) = 24x + 10 \\
38 + 10x = 24x + 10 \\
38 - 10 = 24x - 10x \\
28 = 14x \Rightarrow x = \boxed{2}
$$

✔ Answer: $ \boxed{x = 2} $

---

Problem 6:

Given:
$ m\angle LMN = 108^\circ $
$ m\angle NML = x + 4 $, $ m\angle LMM = 5x - 2 $? Wait — look at diagram.

Diagram:
- Point M, with rays to N, H, and L
- $ \angle LMN = 108^\circ $
- Ray MH splits $ \angle LMN $ into two parts:
- $ \angle NMH = x + 4 $
- $ \angle HML = 5x - 2 $

So:
$$
(x + 4) + (5x - 2) = 108 \\
6x + 2 = 108 \\
6x = 106 \Rightarrow x = \frac{106}{6} = \frac{53}{3} \approx 17.67
$$

Then:
- $ \angle NMH = x + 4 = \frac{53}{3} + 4 = \frac{53 + 12}{3} = \frac{65}{3} \approx 21.67^\circ $
- $ \angle HML = 5x - 2 = 5(\frac{53}{3}) - 2 = \frac{265}{3} - \frac{6}{3} = \frac{259}{3} \approx 86.33^\circ $

Sum: $ 21.67 + 86.33 = 108 $ → correct.

But again, fractional. Maybe typo?

Wait — is $ \angle HML = 5x - 2 $, and $ \angle NMH = x + 4 $, and they add to 108?

Yes.

So:
$$
x + 4 + 5x - 2 = 108 \Rightarrow 6x + 2 = 108 \Rightarrow 6x = 106 \Rightarrow x = \frac{53}{3}
$$

So the two smaller angles:
- $ x + 4 = \frac{53}{3} + \frac{12}{3} = \frac{65}{3} \approx 21.67^\circ $
- $ 5x - 2 = \frac{265}{3} - \frac{6}{3} = \frac{259}{3} \approx 86.33^\circ $

But since the question asks for the measure of the two smaller angles, and both are less than 108, but one is much bigger.

So the two angles are:
- $ \boxed{\frac{65}{3}^\circ} $ and $ \boxed{\frac{259}{3}^\circ} $

Or as decimals: $ \boxed{21.\overline{6}^\circ} $ and $ \boxed{86.\overline{3}^\circ} $

But likely expected as fractions.

✔ Answer: $ \boxed{\frac{65}{3}^\circ} $ and $ \boxed{\frac{259}{3}^\circ} $

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Problem 7:

Given:
$ m\angle ADB = 39^\circ $, $ m\angle ADC = 8x + 5 $, $ m\angle BDC = 3x - 4 $
Find $ m\angle ADC $

From diagram:
- Point D, with rays to A, B, C
- Ray DB lies between DA and DC
- So $ \angle ADC = \angle ADB + \angle BDC $

So:
$$
\angle ADC = \angle ADB + \angle BDC \\
8x + 5 = 39 + (3x - 4) \\
8x + 5 = 35 + 3x \\
8x - 3x = 35 - 5 \\
5x = 30 \Rightarrow x = 6
$$

Now plug into $ m\angle ADC = 8x + 5 = 8(6) + 5 = 48 + 5 = \boxed{53^\circ} $

✔ Answer: $ \boxed{53^\circ} $

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Problem 8:

Given:
$ m\angle RSU = 10x - 9 $
Find degree measures of all 3 angles.

From diagram:
- Ray ST splits $ \angle RSU $ into:
- $ \angle RST = 2x $
- $ \angle TSU = 2x + 75 $

So:
$$
\angle RSU = \angle RST + \angle TSU \\
10x - 9 = 2x + (2x + 75) \\
10x - 9 = 4x + 75 \\
10x - 4x = 75 + 9 \\
6x = 84 \Rightarrow x = 14
$$

Now compute each angle:
- $ \angle RST = 2x = 2(14) = \boxed{28^\circ} $
- $ \angle TSU = 2x + 75 = 28 + 75 = \boxed{103^\circ} $
- $ \angle RSU = 10x - 9 = 140 - 9 = \boxed{131^\circ} $

Check: $ 28 + 103 = 131 $ → correct.

✔ Answer:
- $ \angle RST = 28^\circ $
- $ \angle TSU = 103^\circ $
- $ \angle RSU = 131^\circ $

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $ \boxed{63^\circ} $ |
| 2 | $ \boxed{x = 18} $ |
| 3 | $ \boxed{36^\circ} $ |
| 4 | $ \boxed{x = \dfrac{60}{11}} $ |
| 5 | $ \boxed{x = 2} $ |
| 6 | $ \boxed{\dfrac{65}{3}^\circ} $ and $ \boxed{\dfrac{259}{3}^\circ} $ |
| 7 | $ \boxed{53^\circ} $ |
| 8 | $ \angle RST = 28^\circ $, $ \angle TSU = 103^\circ $, $ \angle RSU = 131^\circ $ |

Let me know if you'd like these formatted neatly or explained further!