Angle Addition Postulate practice worksheet with eight problems and diagrams.

Let’s solve each problem one by one using the Angle Addition Postulate, which states:

> If point B lies in the interior of ∠AOC, then
> m∠AOB + m∠BOC = m∠AOC

This means when two adjacent angles share a common side and vertex, their measures add up to the measure of the larger angle they form together.

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Problem 1

Given:
m∠RUS = 32°, m∠SUT = 31°
Find: m∠TUR

Looking at the diagram: Rays UR, US, UT all emanate from U. Angle TUR is the *entire* angle from T to R, passing through S. So:

> m∠TUR = m∠TUS + m∠SUR
But note: m∠TUS = m∠SUT = 31° (same angle)
and m∠SUR = m∠RUS = 32° (same angle)

So:
> m∠TUR = 31° + 32° = 63°

✔ Answer: 63°

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Problem 2

Given:
m∠GJI = 103°, m∠GJH = 2x + 6, m∠IJH = 3x + 7
Find: x

From the diagram: Ray JH is between rays JG and JI. So:

> m∠GJH + m∠IJH = m∠GJI
→ (2x + 6) + (3x + 7) = 103
→ 5x + 13 = 103
→ 5x = 90
→ x = 18

✔ Answer: x = 18

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Problem 3

Given:
m∠ABC = 87°, m∠ABD = 5x - 9, m∠CBD = 3x
Find: m∠CBD

From diagram: Point D is inside ∠ABC, so:

> m∠ABD + m∠CBD = m∠ABC
→ (5x - 9) + 3x = 87
→ 8x - 9 = 87
→ 8x = 96
→ x = 12

Now find m∠CBD = 3x = 3 × 12 = 36°

✔ Answer: 36°

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Problem 4

Given:
m∠SQR = 52°, m∠PQS = 3x, m∠RQP = 8x - 8
Find: x

Note: The naming might be tricky. Angles ∠PQS and ∠RQP are parts of ∠SQR? Let’s interpret carefully.

Actually, looking at the diagram: Rays QP, QS, QR — likely ray QS is between QP and QR? But the angle given is ∠SQR = 52°, which is from S to R.

Wait — if we assume that ray QP is on one side, QR on the other, and QS is between them, then:

> m∠PQS + m∠SQR = m∠PQR? Not quite.

Actually, the angle names suggest:
- ∠PQS is between P and S
- ∠RQP is between R and P — but that’s the same as ∠PQR?

Let’s re-read: “m∠SQR = 52°”, “m∠PQS = 3x”, “m∠RQP = 8x - 8”

Notice: ∠RQP is the same as ∠PQR — angle at Q between R and P.

If points are arranged P-S-R around Q, then:

> m∠PQS + m∠SQR = m∠PQR → 3x + 52 = 8x - 8

Yes! That makes sense.

So:
> 3x + 52 = 8x - 8
→ 52 + 8 = 8x - 3x
→ 60 = 5x
→ x = 12

✔ Answer: x = 12

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Problem 5

Given:
m∠DEG = 46°, m∠DEF = 24x + 10, m∠GEF = 10x - 8
Find: x

From diagram: Ray EG is between ED and EF. So:

> m∠DEG + m∠GEF = m∠DEF
→ 46 + (10x - 8) = 24x + 10
→ 38 + 10x = 24x + 10
→ 38 - 10 = 24x - 10x
→ 28 = 14x
→ x = 2

✔ Answer: x = 2

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Problem 6

Given:
m∠LMN = 108°, and two smaller angles: (x + 4) and (5x - 2)
Find: Measure of the two smaller angles.

From diagram: Ray MH is between ML and MN. So:

> (x + 4) + (5x - 2) = 108
→ 6x + 2 = 108
→ 6x = 106
→ x = 106/6 = 53/3 ≈ 17.666...

Wait — let’s check:
x + 4 = 53/3 + 12/3 = 65/3 ≈ 21.67°
5x - 2 = 265/3 - 6/3 = 259/3 ≈ 86.33°
Sum = 65/3 + 259/3 = 324/3 = 108° ✔

But maybe we should leave as fractions or decimals? Since it’s a practice sheet, likely expects exact values.

So:

> Smaller angle: x + 4 = 65/3° or 21 2/3°
> Larger small angle: 5x - 2 = 259/3° or 86 1/3°

But since the question says “find the measure of the two smaller angles” — both are “smaller” than the total 108°, so both are answers.

✔ Answer: 65/3° and 259/3° (or 21⅔° and 86⅓°)

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Problem 7

Given:
m∠ADB = 39°, m∠ADC = 8x + 5, m∠BDC = 3x - 4
Find: m∠ADC

From diagram: Point B is inside ∠ADC, so:

> m∠ADB + m∠BDC = m∠ADC
→ 39 + (3x - 4) = 8x + 5
→ 35 + 3x = 8x + 5
→ 35 - 5 = 8x - 3x
→ 30 = 5x
→ x = 6

Now find m∠ADC = 8x + 5 = 8×6 + 5 = 48 + 5 = 53°

✔ Answer: 53°

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Problem 8

Given:
m∠RSU = 10x - 9, and two smaller angles: 2x and 2x + 75
Find: Degree measure of all 3 angles.

From diagram: Ray ST is between SR and SU. So:

> 2x + (2x + 75) = 10x - 9
→ 4x + 75 = 10x - 9
→ 75 + 9 = 10x - 4x
→ 84 = 6x
→ x = 14

Now compute each angle:

- m∠RST = 2x = 2×14 = 28°
- m∠TSU = 2x + 75 = 28 + 75 = 103°
- m∠RSU = 10x - 9 = 140 - 9 = 131°

Check: 28 + 103 = 131 ✔

✔ Answer: 28°, 103°, 131°

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## ✔ Final Answers Summary:

1. 63°
2. x = 18
3. 36°
4. x = 12
5. x = 2
6. 65/3° and 259/3° (or 21⅔° and 86⅓°)
7. 53°
8. 28°, 103°, 131°

Let me know if you’d like diagrams or step-by-step visuals for any problem!