Grade 6 math worksheet on identifying unknown angles using geometric diagrams and angle measurements.

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 XOY is a straight line (180°), then:

> ∠XOZ = 180° – ∠YOZ = 180° – 90° = 90°

✔ Answer: 90°

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2. ∠ROQ = ?

Given:
- ∠POQ = 105°
- P–O–Q is a straight line → 180°
- So, ∠ROQ is the remaining part of the straight line.

But note: ray OR is drawn upward from O, forming an angle with OQ.

So, ∠POQ = 105°, which means the angle from P to R to Q is 105°.

But we want ∠ROQ — that’s the angle from R to O to Q, which is the same as the angle between R and Q.

Wait — actually, since ∠POQ = 105°, and P–O–Q is a straight line, then the angle between R and Q must be:

> ∠ROQ = 180° – 105° = 75°

✔ Answer: 75°

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3. ∠ACD = ?

Given:
- Line AB is straight → 180°
- ∠DCB = 60°
- We need ∠ACD

Note: Point C is on line AB, and D is above, forming triangle-like shape.

So, ∠ACD + ∠DCB = 180° (straight line)

> ∠ACD = 180° – 60° = 120°

✔ Answer: 120°

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4. ∠ACD = ?

Given:
- Line AB is straight → 180°
- ∠DCB = 30°
- So, ∠ACD = ?

Again, ∠ACD + ∠DCB = 180°

> ∠ACD = 180° – 30° = 150°

✔ Answer: 150°

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5. ∠XQP = ?

Given:
- Line XY is straight → 180°
- ∠XQP = 135°? Wait — no, the angle shown is 135° between X–Q and Q–P?

Wait — let's look carefully.

Point Q is on line XY, and ray QP goes upward.

The angle between X–Q and Q–P is marked as 135°.

We are asked for ∠XQP — which is exactly that angle.

But wait — is it asking for the same angle or the adjacent one?

No — ∠XQP is the angle formed by points X–Q–P, so it’s the angle at Q between X and P.

And it's already labeled as 135°.

But that can't be — because if it's already labeled, why ask?

Wait — maybe I misread.

Looking again: The angle between X–Q and Q–P is 135°, and we're to find ∠XQP — which is that angle.

But perhaps the diagram shows 135°, but asks us to confirm?

Wait — no, likely the angle between X–Q and Q–P is 135°, so ∠XQP = 135°.

But that seems too easy.

Alternatively, maybe the angle not shown is what we need?

Wait — no, the label says “∠xqp =” and the diagram shows 135° at that vertex.

But in standard notation, ∠XQP means the angle at Q between X and P.

So yes, ∠XQP = 135°

But let's double-check: Is there a possibility of supplementary?

No — unless they want the other side.

But the diagram shows 135° between X–Q and Q–P.

So, ∠XQP = 135°

✔ Answer: 135°

Wait — this seems odd because usually such questions ask for missing angles.

But here, it's given.

Unless… is the 135° not ∠XQP?

Wait — let's clarify:

- Line XY is horizontal.
- Ray QP goes up-left.
- The angle between X–Q and Q–P is 135° — so yes, that’s ∠XQP.

So yes, it's 135°

✔ Answer: 135°

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6. ∠PTS = ?

Given:
- Line PQ is straight → 180°
- ∠PTQ = 120°
- So, ∠PTS is the angle between P–T–S

But S is going upward from T.

So, the angle between P–T and T–S is 120°, as shown.

So ∠PTS = 120°

Wait — but again, it's already labeled.

But the question asks to identify the unknown angle.

But here, the angle is labeled as 120°.

Is ∠PTS the same as that?

Yes — points P–T–S form the angle.

So ∠PTS = 120°

✔ Answer: 120°

Wait — this seems suspiciously direct.

But maybe the intention is just to read the angle.

Alternatively, could it be asking for the adjacent angle?

No — ∠PTS is clearly the angle shown.

So, 120°

✔ Answer: 120°

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7. ∠XYB = ?

Given:
- Line AB is straight → 180°
- ∠AYX = 15°
- We are to find ∠XYB

Points:
- Y is on AB
- Ray YX goes upward-left
- Angle between A–Y and Y–X is 15°

So, ∠AYX = 15°

Now, ∠XYB is the angle from X–Y–B

So, total straight line at Y: ∠AYX + ∠XYB = 180°

> ∠XYB = 180° – 15° = 165°

✔ Answer: 165°

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8. ∠SOQ = ?

Given:
- Line PQ is straight → 180°
- ∠POS = 100°
- We are to find ∠SOQ

So, from P to O to Q is a straight line.

Ray OS makes a 100° angle with OP.

So, ∠POS = 100°

Then, ∠SOQ is the remaining angle from S to Q.

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 | 135° |
| 6. ∠PTS | 120° |
| 7. ∠XYB | 165° |
| 8. ∠SOQ | 80° |

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🔍 Explanation Summary:

- Straight lines: Angles on a straight line sum to 180°
- Right angles: 90°
- Adjacent angles: Add to 180° if on a straight line
- Always identify which angle is being asked (e.g., ∠XQP means angle at Q between X and P)
- Use subtraction: Unknown = 180° – Given angle (if on straight line)

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