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 → ∠XOY = 180°

So, ∠XOZ is the angle from X to Z through O.

Since ∠YOZ = 90°, then ∠XOZ = 180° – 90° = 90°

✔ Answer: 90°

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2. ∠ROQ = ?

Given:
- ∠POQ = 180° (straight line)
- ∠POR = 105°

So, ∠ROQ = 180° – 105° = 75°

✔ Answer: 75°

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3. ∠ACD = ?

Given:
- ∠DCB = 60°
- AB is a straight line → ∠ACB = 180°

So, ∠ACD = 180° – 60° = 120°

✔ Answer: 120°

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4. ∠ACD = ?

Given:
- ∠DCB = 30°
- AB is a straight line → ∠ACB = 180°

So, ∠ACD = 180° – 30° = 150°

✔ Answer: 150°

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5. ∠XQP = ?

Given:
- ∠XQY = 135°
- PQ is a ray forming an angle with XY

But note: ∠XQP is the angle between XQ and QP.

Since ∠XQY = 135°, and the total around point Q on a straight line is 180°, the remaining angle (∠PQY) would be:

∠PQY = 180° – 135° = 45°

But we want ∠XQP — which is the angle from X to P via Q.

Wait: Point P is above the line, and the angle between XQ and QP is not directly given.

But looking at the diagram:
- Ray QP makes an angle with QY.
- The angle between XQ and QP is the same as the angle adjacent to 135°, so:

∠XQP = 180° – 135° = 45°

✔ Answer: 45°

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6. ∠PTS = ?

Given:
- ∠PTQ = 120°
- PTQ is a straight line? No — actually, P–T–Q is a straight line, and S is a ray from T.

So, ∠PTS is the angle between PT and TS.

But we're told that ∠PTQ = 120° — this is the angle from P to Q via T.

So, ∠PTS = 120° — because it’s the same angle.

Wait — let’s clarify:

If ∠PTQ = 120°, and T is the vertex, then ∠PTS is part of that?

No — actually, if the ray TS forms a 120° angle with TQ, then since P–T–Q is straight, ∠PTS = 180° – 120° = 60°

Wait — no: the angle shown is between TS and TQ, which is 120°.

Then, ∠PTS is the angle from P to T to S.

Since P–T–Q is straight, and ∠STQ = 120°, then ∠PTS = 180° – 120° = 60°

✔ Answer: 60°

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7. ∠XYB = ?

Given:
- ∠XYA = 15°
- AB is a straight line → ∠AYB = 180°

So, ∠XYB = 180° – 15° = 165°

✔ Answer: 165°

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8. ∠SOQ = ?

Given:
- ∠POQ = 180° (straight line)
- ∠POS = 100°

So, ∠SOQ = 180° – 100° = 80°

✔ Answer: 80°

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✔ Final Answers:

| Question | Answer |
|--------|--------|
| 1. ∠XOZ | 90° |
| 2. ∠ROQ | 75° |
| 3. ∠ACD | 120° |
| 4. ∠ACD | 150° |
| 5. ∠XQP | 45° |
| 6. ∠PTS | 60° |
| 7. ∠XYB | 165° |
| 8. ∠SOQ | 80° |

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🔍 Summary of Concepts Used:

- Straight line = 180°: Any angle formed on a straight line adds up to 180°.
- Supplementary angles: Two angles on a straight line sum to 180°.
- Right angle = 90°: Indicated by a square corner.

These are foundational concepts in geometry for Grade 6 students.

Let me know if you'd like this formatted as a printable answer sheet!