Math worksheet for 7th-grade students focusing on calculating unknown angles using linear pair and opposite angle theorems.

Let's solve each problem step by step using the concepts of linear pairs, adjacent angles, and vertically opposite angles.

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🔹 Problem 1:

Figure:
- A straight line AC with point O on it.
- ∠AOB = 60°, ∠BOC = x°
- These two angles form a linear pair (they are adjacent and add up to 180°).

So:
$$
60^\circ + x = 180^\circ \\
x = 180^\circ - 60^\circ = 120^\circ
$$

✔ x = 120

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🔹 Problem 2:

Figure:
- Straight line PR, point O on it.
- ∠POQ = 3x°, ∠QOR = 2x°
- They form a linear pair → sum = 180°

$$
3x + 2x = 180^\circ \\
5x = 180^\circ \\
x = 36^\circ
$$

✔ x = 36

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🔹 Problem 3:

Figure:
- Line LM with point O.
- ∠PON = x°, ∠NOM = 60°, ∠POL = 35°
- ∠PON and ∠NOM are adjacent, but we need to find x.

But note: ∠POL = 35°, ∠PON = x°, and ∠NOM = 60°

Since L-O-M is a straight line, total angle = 180°

So:
$$
∠POL + ∠PON + ∠NOM = 180^\circ \\
35^\circ + x + 60^\circ = 180^\circ \\
x = 180 - 95 = 85^\circ
$$

✔ x = 85

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🔹 Problem 4:

Figure:
- Angles around point O: 3x°, 2x°, x°, and 2x° (wait — check labels)

Wait: angles are:
- ∠ROQ = 2x°
- ∠QOP = x°
- ∠POS = 2x°
- ∠SOR = 3x°

All four angles around point O sum to 360°

So:
$$
3x + 2x + x + 2x = 360^\circ \\
8x = 360^\circ \\
x = 45^\circ
$$

✔ x = 45

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🔹 Problem 5:

Figure:
- Two intersecting lines forming vertically opposite angles.
- One angle = 105°, another = 3x°

Vertically opposite angles are equal.

So:
$$
3x = 105^\circ \\
x = 35^\circ
$$

✔ x = 35

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🔹 Problem 6:

Figure:
- Line AB with point O.
- ∠AOC = (2y + 5)°, ∠COB = 3x°
- Given: x = 25 (already filled in), find y.

We know:
- ∠AOC + ∠COB = 180° (linear pair)
- So:
$$
(2y + 5) + 3x = 180 \\
\text{Substitute } x = 25: \\
(2y + 5) + 3(25) = 180 \\
2y + 5 + 75 = 180 \\
2y + 80 = 180 \\
2y = 100 \\
y = 50
$$

✔ y = 50

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🔹 Problem 7:

Figure:
- Three rays from point O: OA, OB, OC
- ∠AOD = (x+10)°, ∠DOC = x°, ∠COB = (x+20)°
- All three angles form a straight line? Wait — look: D–O–C–B? No, seems like angles at point O along a straight line?

Wait — actually, points A, O, B are on a straight line. So ∠AOB = 180°

Angles:
- ∠AOD = (x+10)°
- ∠DOC = x°
- ∠COB = (x+20)°

They add up to 180°:

$$
(x+10) + x + (x+20) = 180 \\
3x + 30 = 180 \\
3x = 150 \\
x = 50
$$

✔ x = 50

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🔹 Problem 8:

Figure:
- Coordinate plane-like setup, with angles at origin O.
- Angles: ∠COB = 3x°, ∠BOA = 3x°, ∠AOD = x°
- The full circle is 360°, but here it looks like the angles are around a point.

Wait: C–O–A is horizontal, O–B vertical up, O–D down-right.

So:
- ∠COB = 3x° (from negative x-axis to positive y-axis)
- ∠BOA = 3x° (positive y-axis to positive x-axis)
- ∠AOD = x° (positive x-axis to ray OD)
- Then ∠DOC = ? But not labeled.

Wait: Let’s assume these are angles around point O.

But notice: From C to B: 3x°, B to A: 3x°, A to D: x°, and then D back to C?

But there's no direct info. Wait — perhaps only some angles are shown.

Actually, looking carefully:
- ∠COB = 3x°
- ∠BOA = 3x°
- ∠AOD = x°
- And the remaining angle from D to C?

But since it's symmetric, maybe not.

Alternatively, consider that ∠COB and ∠BOA are both 3x°, and they go from C to A via B.

But if C–O–A is a straight line, then ∠COA = 180°

Then:
- ∠COB + ∠BOA = 3x + 3x = 6x = 180°
→ x = 30°

But wait — is that correct?

Yes! Because C–O–A is a straight line. So the angle from C to A through B is 180°.

So:
$$
∠COB + ∠BOA = 3x + 3x = 6x = 180^\circ \\
x = 30^\circ
$$

✔ x = 30

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🔹 Problem 9:

Figure:
- Two lines intersecting at O.
- ∠AOB = 70°, ∠BOC = 2x°
- Also, ∠AOB and ∠BOC are adjacent and form a straight line? Not necessarily.

Wait — points A–O–C seem to be a straight line.

So ∠AOC = 180°

And ∠AOB = 70°, ∠BOC = 2x°

So:
$$
70^\circ + 2x = 180^\circ \\
2x = 110^\circ \\
x = 55^\circ
$$

✔ x = 55

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🔹 Problem 10:

Figure:
- Point O with rays OP, OQ, OR, OS
- ∠POQ = 60°, ∠QOR = 4x°, ∠ROS = 40°
- Are all these angles around point O?

Wait — check: Is there a straight line?

No — but let’s see if any are supplementary.

Wait: Look at the directions — seems like angles around point O.

But do they form a straight line?

Wait — look: ∠POQ = 60°, ∠QOR = 4x°, ∠ROS = 40°, and also ∠SOP?

But no — maybe only three angles are given.

Wait — perhaps P–O–S is a straight line?

Check: ∠POQ = 60°, ∠QOR = 4x°, ∠ROS = 40°

If P–O–S is a straight line, then total angle = 180°

So:
$$
∠POQ + ∠QOR + ∠ROS = 180^\circ \\
60 + 4x + 40 = 180 \\
4x + 100 = 180 \\
4x = 80 \\
x = 20
$$

✔ x = 20

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## ✔ Final Answers:

| Problem | Answer |
|--------|--------|
| 1 | x = 120 |
| 2 | x = 36 |
| 3 | x = 85 |
| 4 | x = 45 |
| 5 | x = 35 |
| 6 | x = 25, y = 50 |
| 7 | x = 50 |
| 8 | x = 30 |
| 9 | x = 55 |
| 10 | x = 20 |

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