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.

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

✔ Answer: $ \boxed{18} $

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

Given:
- $ m\angle ABC = 87^\circ $
- $ m\angle CBD = 3x $
- $ m\angle ABD = 5x - 9 $
- And $ \angle ABC $ is made up of $ \angle CBD $ and $ \angle ABD $? Wait — let’s look at the diagram.

Points: B is vertex. Ray BC, BD, BA.
- $ \angle CBD = 3x $
- $ \angle DBA = 5x - 9 $
- So total $ \angle ABC = \angle CBD + \angle DBA $

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 $.

Wait — $ \angle PQS $ and $ \angle RQP $ share ray $ QP $, but the diagram shows rays from $ Q $: P, S, R.

From the diagram:
- $ \angle SQR $ is composed of $ \angle SQP $ and $ \angle PQR $
- But given: $ m\angle PQS = 3x $, $ m\angle RQP = 8x - 8 $

Note: $ \angle PQS $ is same as $ \angle SQP $, so they add to form $ \angle SQR $:
$$
m\angle SQP + m\angle PQR = m\angle SQR
\Rightarrow 3x + (8x - 8) = 52
$$
$$
11x - 8 = 52
\Rightarrow 11x = 60
\Rightarrow x = \frac{60}{11} \approx 5.45
$$

But wait — check again. Is this correct?

Wait — $ \angle PQS $ and $ \angle RQP $ — both start from $ Q $, but do they form $ \angle SQR $? Let’s think:

- $ \angle PQS $: from P to S
- $ \angle RQP $: from R to P? That doesn't make sense.

Actually, the order matters.

Looking at the diagram:
- Rays: QP, QS, QR
- So $ \angle SQR $ is between S and R
- Ray QS is inside $ \angle SQR $? No — actually, QS is likely between QP and QR.

Wait — better interpretation:

From diagram:
- Points: P — Q — S — R (around a point)
- So $ \angle SQR $ is from S to R through Q
- But $ \angle PQS $ is from P to S, $ \angle RQP $ is from R to P? That can’t be.

Wait — probably it's:
- $ \angle PQS $ and $ \angle SQR $ together make $ \angle PQR $? But no.

Wait — perhaps $ \angle SQR $ is split into $ \angle SQP $ and $ \angle PQR $?

But the labels say:
- $ m\angle PQS = 3x $
- $ m\angle RQP = 8x - 8 $

But $ \angle RQP $ is the same as $ \angle PQR $, and $ \angle PQS $ is $ \angle SQP $

So $ \angle SQR = \angle SQP + \angle PQR = 3x + (8x - 8) = 11x - 8 $

Set equal to $ 52^\circ $:
$$
11x - 8 = 52
\Rightarrow 11x = 60
\Rightarrow x = \frac{60}{11}
$$

That’s not a whole number. Maybe mistake in interpretation.

Wait — maybe $ \angle SQR $ is the large angle, and $ \angle PQS $ and $ \angle RQP $ are parts of it?

But $ \angle PQS $ is from P to S, $ \angle RQP $ from R to P — they might overlap.

Alternative idea: Perhaps ray QS lies between QP and QR, so:
- $ \angle PQS = 3x $
- $ \angle SQR = 52^\circ $
- Then $ \angle PQR = \angle PQS + \angle SQR = 3x + 52^\circ $? But we're told $ m\angle RQP = 8x - 8 $

Wait — $ \angle RQP $ is the same as $ \angle PQR $, so:
$$
\angle PQR = \angle PQS + \angle SQR
\Rightarrow 8x - 8 = 3x + 52
$$
$$
8x - 8 = 3x + 52
\Rightarrow 5x = 60
\Rightarrow x = 12
$$

Yes! That makes more sense.

So $ \angle PQR = 8x - 8 = 8(12) - 8 = 96 - 8 = 88^\circ $

And $ \angle PQS = 3x = 36^\circ $, $ \angle SQR = 52^\circ $, so total $ 36 + 52 = 88^\circ $ → checks out.

✔ Answer: $ \boxed{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:
- Point E, rays ED, EG, EF
- $ \angle DEG $ is between D and G
- $ \angle DEF $ is from D to F
- $ \angle GEF $ is from G to F

So $ \angle DEF = \angle DEG + \angle GEF $

So:
$$
24x + 10 = 46 + (10x - 8)
$$
$$
24x + 10 = 46 + 10x - 8 = 10x + 38
$$
$$
24x - 10x = 38 - 10
\Rightarrow 14x = 28
\Rightarrow x = 2
$$

Check:
- $ \angle DEF = 24(2) + 10 = 48 + 10 = 58^\circ $
- $ \angle GEF = 10(2) - 8 = 20 - 8 = 12^\circ $
- $ \angle DEG = 46^\circ $
- $ 46 + 12 = 58 $ → Correct!

✔ Answer: $ \boxed{2} $

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

Given:
- $ m\angle LMN = 108^\circ $
- $ m\angle LMP = 5x - 2 $
- $ m\angle PMN = x + 14 $

Assuming point P is on $ \angle LMN $, so:
$$
m\angle LMN = m\angle LMP + m\angle PMN
\Rightarrow (5x - 2) + (x + 14) = 108
\Rightarrow 6x + 12 = 108
\Rightarrow 6x = 96
\Rightarrow x = 16
$$

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

So the two smaller angles are $ 78^\circ $ and $ 30^\circ $

But "smaller" — both are smaller than 108°, but only one is smaller than the other.

Wait — “the two smaller angles” — but there are only two parts.

So just report them.

✔ Answer: $ \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:
- Point D, rays DA, DB, DC
- $ \angle ADC $ is from A to C
- $ \angle ADB $ is from A to B
- $ \angle BDC $ is from B to C

So $ \angle ADC = \angle ADB + \angle BDC $

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

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

✔ Answer: $ \boxed{53^\circ} $

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

Given:
- $ m\angle RST = 2x $
- $ m\angle TSU = 2x + 75 $
- $ m\angle RSU = 10x - 9 $

From diagram:
- $ \angle RSU = \angle RST + \angle TSU $

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

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

Check: $ 28 + 103 = 131 $ → Correct!

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

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✔ Final Answers Summary:

1. $ \boxed{63^\circ} $
2. $ \boxed{18} $
3. $ \boxed{36^\circ} $
4. $ \boxed{12} $
5. $ \boxed{2} $
6. $ \boxed{78^\circ} $ and $ \boxed{30^\circ} $
7. $ \boxed{53^\circ} $
8. $ \angle RST = \boxed{28^\circ}, \angle TSU = \boxed{103^\circ}, \angle RSU = \boxed{131^\circ} $

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