Identifying Unknown Angles Worksheets | Geometry - Free Printable

Let's solve each of the problems on the worksheet step by step. The goal is to identify the unknown angles using basic geometry principles such as:

- Straight line = 180°
- Right angle = 90°
- Angles around a point = 360°
- Supplementary angles add up to 180°
- Complementary angles add up to 90°

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1. ∠XOZ = ?

We are given:
- A right angle at O between Y and Z → ∠YOZ = 90°
- X–O–Y is a straight line → 180°

So, ∠XOZ is the angle from X to Z.

Since ∠YOZ = 90°, and X–O–Y is a straight line, then:
- ∠XOZ = 180° – ∠YOZ = 180° – 90° = 90°

✔ ∠XOZ = 90°

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2. ∠ROQ = ?

Given:
- ∠POQ = 105° (angle between p and q)
- Ray OR is drawn upward from point O
- We need ∠ROQ (from R to Q)

But we are not told if this is a straight line or if there’s more context. However, since the angle shown is 105° between p and r, and p–o–q is a straight line (180°), then:

- ∠ROQ = 180° – 105° = 75°

✔ ∠ROQ = 75°

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3. ∠ACD = ?

Given:
- Point C lies on line ab
- CD makes a 60° angle with cb
- So ∠DCB = 60°

We want ∠ACD — the angle between AC and CD.

Since ab is a straight line: ∠ACB = 180°

Then:
- ∠ACD = 180° – ∠DCB = 180° – 60° = 120°

✔ ∠ACD = 120°

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4. ∠ACD = ?

Given:
- Line ab, point C
- CD makes 30° with cb → ∠DCB = 30°
- Want ∠ACD

Again, ab is straight → ∠ACB = 180°

So:
- ∠ACD = 180° – 30° = 150°

✔ ∠ACD = 150°

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5. ∠XQP = ?

Given:
- Line XY, point Q
- Angle between XQ and PQ is 135°
- We are to find ∠XQP → that is the angle between X and P at Q

Wait — but it's already labeled as 135°? Let's clarify.

Actually, the diagram shows a ray QP forming a 135° angle with QX.

So ∠XQP is the angle at Q between X and P → which is 135°

But wait — let’s confirm: Is it asking for the same angle?

Yes, ∠XQP is the angle formed by rays QX and QP → and it's marked as 135°

✔ ∠XQP = 135°

(No calculation needed; it's directly given.)

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6. ∠PTS = ?

Given:
- Line pq, point T
- Ray TS forms 120° with TQ
- So ∠QT S = 120°
- Want ∠PTS → angle between PT and ST

But pt is in the opposite direction of qt.

So, ∠PTQ = 180° (straight line), and ∠QTS = 120°

Then:
- ∠PTS = 180° – 120° = 60°

✔ ∠PTS = 60°

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7. ∠XYB = ?

Given:
- Line ab, point Y
- Ray YX forms 15° with Ya
- So ∠AYX = 15°
- Want ∠XYB → angle between X and B at Y

Since ab is straight → ∠AYB = 180°

Then:
- ∠XYB = 180° – ∠AYX = 180° – 15° = 165°

✔ ∠XYB = 165°

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8. ∠SOQ = ?

Given:
- Line pq, point O
- Ray OS forms 100° with OP
- So ∠POS = 100°
- Want ∠SOQ → angle between SO and OQ

Since pq is straight → ∠POQ = 180°

So:
- ∠SOQ = 180° – 100° = 80°

✔ ∠SOQ = 80°

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✔ Final Answers:

| Question | Answer |
|--------|--------|
| 1. ∠XOZ | 90° |
| 2. ∠ROQ | 75° |
| 3. ∠ACD | 120° |
| 4. ∠ACD | 150° |
| 5. ∠XQP | 135° |
| 6. ∠PTS | 60° |
| 7. ∠XYB | 165° |
| 8. ∠SOQ | 80° |

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🔍 Summary of Concepts Used:

- Straight lines: Angles on a straight line sum to 180°
- Supplementary angles: Two angles that add to 180°
- Right angles: Always 90°
- Angle labeling: Carefully identify the vertex and rays involved

Let me know if you'd like a visual explanation or printable version!