Angle Addition Postulate practice worksheet with eight problems and diagrams.

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.

---

1. If $ m\angle RUS = 32^\circ $ and $ m\angle SUT = 31^\circ $, what is $ m\angle TUR $?

From the diagram:
- $ \angle RUS $ and $ \angle SUT $ are adjacent angles forming $ \angle RUT $.
- So:
$$
m\angle RUT = m\angle RUS + m\angle SUT = 32^\circ + 31^\circ = 63^\circ
$$

✔ Answer: $ \boxed{63^\circ} $

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2. $ 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 $ form $ \angle GJI $
- So:
$$
(2x + 6) + (3x + 7) = 103
$$
$$
5x + 13 = 103
$$
$$
5x = 90 \Rightarrow x = 18
$$

✔ Answer: $ \boxed{x = 18} $

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3. Find $ m\angle CBD $ if $ m\angle ABC = 87^\circ $.

Given:
- $ m\angle ABC = 87^\circ $
- $ \angle ABC $ is split into $ \angle CBD $ and $ \angle DBA $
- $ m\angle CBD = 3x $
- $ m\angle DBA = 5x - 9 $

So:
$$
3x + (5x - 9) = 87
$$
$$
8x - 9 = 87
$$
$$
8x = 96 \Rightarrow x = 12
$$

Now find $ m\angle CBD = 3x = 3(12) = 36^\circ $

✔ Answer: $ \boxed{36^\circ} $

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

Note: The angles $ \angle PQS $ and $ \angle RQP $ together make up $ \angle PQR $, but we're given $ \angle SQR = 52^\circ $, so let’s look at the diagram carefully.

Wait — from the diagram:
- Point $ Q $ has rays to $ P $, $ S $, and $ R $
- $ \angle SQR = 52^\circ $
- $ \angle PQS = 3x $
- $ \angle RQP = 8x - 8 $

But $ \angle PQS $ and $ \angle SQR $ are adjacent angles forming $ \angle PQR $? Wait — actually, it looks like $ \angle PQS $ and $ \angle SQR $ are parts of $ \angle PQR $, with $ S $ between $ P $ and $ R $.

So:
$$
m\angle PQS + m\angle SQR = m\angle PQR
$$

But we are told $ m\angle SQR = 52^\circ $, and $ m\angle PQS = 3x $, $ m\angle RQP = 8x - 8 $

Wait — $ m\angle RQP $ is the same as $ m\angle PQR $? That seems conflicting.

Actually, rechecking: $ \angle RQP $ is the same as $ \angle PQR $, but we’re told $ m\angle RQP = 8x - 8 $. But $ \angle SQR = 52^\circ $ is part of $ \angle PQR $.

So likely: $ \angle PQS $ and $ \angle SQR $ add up to $ \angle PQR $

So:
$$
m\angle PQS + m\angle SQR = m\angle PQR
\Rightarrow 3x + 52 = 8x - 8
$$

Now solve:
$$
3x + 52 = 8x - 8
\Rightarrow 52 + 8 = 8x - 3x
\Rightarrow 60 = 5x \Rightarrow x = 12
$$

✔ Answer: $ \boxed{x = 12} $

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

From the diagram:
- Ray $ EG $ splits $ \angle DEF $ into $ \angle DEG $ and $ \angle GEF $
- So:
$$
m\angle DEG + m\angle GEF = m\angle DEF
$$
$$
46 + (10x - 8) = 24x + 10
$$
$$
38 + 10x = 24x + 10
$$
$$
38 - 10 = 24x - 10x
\Rightarrow 28 = 14x \Rightarrow x = 2
$$

✔ Answer: $ \boxed{x = 2} $

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6. $ m\angle LMN = 108^\circ $. Find the measures of the two smaller angles.

From the diagram:
- $ \angle LMN $ is split into $ \angle LMM $ and $ \angle NMM $? No — ray $ MH $ splits $ \angle LMN $
- So: $ m\angle LMN = m\angle LMH + m\angle HMN $
- Given: $ m\angle LMH = 5x - 2 $, $ m\angle HMN = x + 14 $

So:
$$
(5x - 2) + (x + 14) = 108
\Rightarrow 6x + 12 = 108
\Rightarrow 6x = 96 \Rightarrow x = 16
$$

Now compute:
- $ m\angle LMH = 5(16) - 2 = 80 - 2 = 78^\circ $
- $ m\angle HMN = 16 + 14 = 30^\circ $

✔ Answer: The two smaller angles are $ \boxed{78^\circ} $ and $ \boxed{30^\circ} $

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

From the diagram:
- $ \angle ADC $ is made up of $ \angle ADB $ and $ \angle BDC $
- So:
$$
m\angle ADB + m\angle BDC = m\angle ADC
$$
$$
39 + (3x - 4) = 8x + 5
$$
$$
35 + 3x = 8x + 5
$$
$$
35 - 5 = 8x - 3x \Rightarrow 30 = 5x \Rightarrow x = 6
$$

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

✔ Answer: $ \boxed{53^\circ} $

---

8. $ m\angle RSU = 10x - 9 $. Find the degree measure of all 3 angles.

From the diagram:
- $ \angle RSU $ is split into $ \angle RST $ and $ \angle TSU $
- $ m\angle RST = 2x $
- $ m\angle TSU = 2x + 75 $
- And $ m\angle RSU = 10x - 9 $

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

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

Check: $ 28 + 103 = 131 $ ✔

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

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $ 63^\circ $ |
| 2 | $ x = 18 $ |
| 3 | $ 36^\circ $ |
| 4 | $ x = 12 $ |
| 5 | $ x = 2 $ |
| 6 | $ 78^\circ $ and $ 30^\circ $ |
| 7 | $ 53^\circ $ |
| 8 | $ 28^\circ, 103^\circ, 131^\circ $ |

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