Linear Pairs of Angles Worksheets - Free Printable

Since I can't view or access images directly, I’ll help you solve the problem based on the description of the worksheet titled "Linear Pairs". This worksheet typically involves identifying linear pairs of angles — two adjacent angles that form a straight line (i.e., their non-common sides are opposite rays and their measures add up to 180°).

Let’s go through each part step by step.

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Part A: Identify all the linear pairs in each figure

A linear pair consists of two adjacent angles whose non-common sides form a straight line (i.e., they are supplementary and share a common vertex and side).

We’ll assume the figures are standard configurations involving intersecting lines or rays forming angles at a point.

#### Figure 1)
Assume this is two lines intersecting at point O, forming four angles: ∠AOB, ∠BOC, ∠COD, ∠DOA.

- Linear pairs:
- ∠AOB and ∠BOC (if they form a straight line)
- ∠BOC and ∠COD
- ∠COD and ∠DOA
- ∠DOA and ∠AOB

But actually, only adjacent angles that form a straight line count. So if lines AB and CD intersect at O, then:

- ∠AOB and ∠BOC → no, unless B is on the straight line.
Wait — better to assume standard notation.

Let’s suppose lines AC and BD cross at O.

Then the linear pairs are:
- ∠AOB and ∠BOC → not necessarily linear unless A-O-C is straight.
Actually, correct approach:

When two lines intersect at a point, they form two pairs of vertical angles, and four linear pairs.

Each angle has two possible linear pairs with its adjacent angles.

For example, if two lines cross at point O, forming angles ∠1, ∠2, ∠3, ∠4 around the point:

- ∠1 and ∠2 → linear pair
- ∠2 and ∠3 → linear pair
- ∠3 and ∠4 → linear pair
- ∠4 and ∠1 → linear pair

So, four linear pairs total.

But without seeing the image, let's assume standard configurations.

Let’s now proceed with typical answers for such problems.

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Common Assumptions for Figures

#### 1) Two lines crossing at a point (X-shaped)

Angles labeled: ∠1, ∠2, ∠3, ∠4 around point O.

Linear pairs:
- ∠1 and ∠2
- ∠2 and ∠3
- ∠3 and ∠4
- ∠4 and ∠1

(If angles are labeled clockwise around the point.)

So answer:
∠1 & ∠2, ∠2 & ∠3, ∠3 & ∠4, ∠4 & ∠1

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#### 2) Three rays from a point (like a "Y" shape)

Suppose rays OA, OB, OC from point O.

Then:
- ∠AOB and ∠BOC → may be linear if A-O-C is straight
- But likely, one ray is opposite another.

More commonly, it's two lines intersecting, or three rays forming angles.

Assuming two lines intersecting again, same as above.

Alternatively, if it's a single line with a ray coming out, like:

- Line ABC, with ray OD from O.

Then linear pairs would be:
- ∠AOD and ∠DOC (if D is off the line), but only if they make a straight line.

Without image, we need to assume.

But let’s move to Part B, which is clearer.

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Part B: Identify the angles that make a linear pair with each specified angle

This part shows a star-like figure: multiple rays from a central point O, forming several angles.

Typical setup: Three lines intersecting at a point, creating six angles around the point.

Labeling: Rays OA, OB, OC, OD, OE, OF going around the circle.

But more commonly, it’s three lines through a point, so 6 rays.

Let’s assume the diagram has six rays from point O, forming angles like:

- ∠AOB, ∠BOC, ∠COD, ∠DOE, ∠EOF, ∠FOA

And some are straight lines.

But usually, in such problems, some rays are opposite, forming straight lines.

Let’s assume the following configuration (standard):

- Ray OA and ray OD are opposite (straight line)
- Ray OB and ray OE are opposite
- Ray OC and ray OF are opposite

So three lines passing through O: AOD, BOE, COF

Now, angles formed:

- ∠AOB, ∠BOC, ∠COD, ∠DOE, ∠EOF, ∠FOA

Now, a linear pair means two adjacent angles that together form a straight line.

So:

#### 1) ∠AOB
- Adjacent angles: ∠BOC and ∠FOA
- But only one of them will form a straight line with ∠AOB.

Which angle, when added to ∠AOB, makes a straight line?

If OA and OD are opposite, and OB is between OA and OC, then:

- The straight line is AOD → so any angle from OA to OD forms a straight line.

So ∠AOB + ∠BOD = 180° → but ∠BOD is not a single angle.

Wait — we need adjacent angles.

Better: If ray OB is between OA and OC, and OA and OD are opposite, then:

The angle adjacent to ∠AOB on the other side is ∠BOC, but that doesn’t make a straight line unless C is on the opposite side.

Alternative: Assume the rays are arranged symmetrically.

Let’s use a common labeling:

Suppose rays are: OA, OB, OC, OD, OE, OF going clockwise around O.

Assume:
- OA and OD are opposite (so AOD is a straight line)
- OB and OE are opposite
- OC and OF are opposite

Then, for each angle, the linear pair is the adjacent angle on the other side of the straight line.

#### Now, solve each:

##### 1) ∠AOB

- ∠AOB is between OA and OB
- The straight line through O is AOD → so the angle on the other side of OA is ∠FOA? Wait.

Wait: if OA and OD are opposite, then the straight line is AOD.

So any angle starting from OA and going toward OD will sum to 180°.

But ∠AOB is small.

To form a linear pair with ∠AOB, we need an angle adjacent to it that completes the straight line.

But since OB is not on the opposite ray, the only way is if the next ray after OB is OC, etc.

Wait — perhaps better to think: the angle adjacent to ∠AOB along the same straight line.

But ∠AOB is not on a straight line unless OB is on the opposite ray.

Ah — here's the key: a linear pair must be adjacent and form a straight line.

So for ∠AOB, the only possible linear pair is the angle that shares the side OB and goes in the opposite direction from OA.

But since OA and OD are opposite, the angle that is adjacent to ∠AOB and lies on the straight line AOD is ∠BOD — but ∠BOD is not a single angle unless it's broken.

Wait — unless there is a ray between B and D.

But if rays are OA, OB, OC, OD, then:

- ∠AOB and ∠BOD are adjacent? Only if B and D are consecutive.

No — better: suppose the rays are labeled around the circle: A, B, C, D, E, F.

And suppose A and D are opposite, B and E, C and F.

Then:

- ∠AOB is between OA and OB
- The angle adjacent to it on the other side is ∠BOC
- But neither ∠BOC nor ∠AOB is on a straight line unless OB is on a straight line.

Wait — only when two rays are opposite do they form a straight line.

So for ∠AOB, the linear pair would be the angle that, together with ∠AOB, makes a straight line.

That happens only if the two angles are adjacent and their non-common sides are opposite rays.

So for ∠AOB, the linear pair would be the angle that shares side OB and has the other side on the opposite ray from OA.

But OA’s opposite is OD.

So if there is a ray OD, then the angle between OB and OD might include ∠BOD.

But unless there’s a ray between B and D, we can’t say.

Alternatively, maybe the figure has only three lines through O, so six angles.

Let’s define:

- Line 1: OA and OD (opposite)
- Line 2: OB and OE (opposite)
- Line 3: OC and OF (opposite)

Then angles:
- ∠AOB, ∠BOC, ∠COD, ∠DOE, ∠EOF, ∠FOA

Now, look at ∠AOB:

- It is between OA and OB
- Its adjacent angles are ∠BOC and ∠FOA
- But to form a linear pair, we need an angle adjacent to it that completes a straight line.

But OA and OD are opposite, so the straight line is AOD.

So any angle from OA to OD, passing through B and C, would sum to 180°.

But ∠AOB is only part of it.

So the angle adjacent to ∠AOB that is on the straight line AOD is ∠BOD — but that’s not a single labeled angle.

Unless the figure has only two rays per line.

Wait — perhaps the figure is simpler: two lines intersecting, forming four angles.

But Part B shows a star with many rays.

Another possibility: three lines through a point, so six angles.

But let’s take a common textbook example.

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Standard Answer Pattern for Part B

In most worksheets like this, the diagram has three lines intersecting at a point, creating six angles.

Let’s assume the rays are labeled: A, B, C, D, E, F around point O.

With:
- A and D opposite
- B and E opposite
- C and F opposite

Then the angles are:
- ∠AOB, ∠BOC, ∠COD, ∠DOE, ∠EOF, ∠FOA

Now, for each given angle, find the angle that forms a linear pair.

Recall: two angles form a linear pair if they are adjacent and their non-common sides are opposite rays.

So:

#### 1) ∠AOB

- Adjacent angles: ∠BOC and ∠FOA
- But which one forms a straight line?

Look at side OA: its opposite is OD.
Side OB: its opposite is OE.

So the angle adjacent to ∠AOB that uses the opposite ray of OA is not directly available.

Wait — the linear pair for ∠AOB would be the angle that shares side OB and has the other side on the opposite ray of OA.

But OA’s opposite is OD.

So the angle between OB and OD is ∠BOD, which is ∠BOC + ∠COD.

But that’s not a single angle.

So unless there is a ray between B and D, it’s not a single angle.

Therefore, the only way is if ∠AOB and ∠BOD are adjacent and form a straight line — but ∠BOD is not a single labeled angle.

Wait — perhaps the figure has only two lines.

Let’s assume two lines intersecting at O, forming four angles: ∠1, ∠2, ∠3, ∠4.

Then linear pairs are:
- ∠1 and ∠2
- ∠2 and ∠3
- ∠3 and ∠4
- ∠4 and ∠1

But Part B has specific names: ∠AOB, ∠BOC, etc.

So likely, the rays are labeled.

Let’s suppose the rays are: OA, OB, OC, OD, with OA and OC opposite, OB and OD opposite.

Then angles:
- ∠AOB, ∠BOC, ∠COD, ∠DOA

But ∠AOB and ∠BOC are adjacent, but not necessarily linear unless AOB is straight.

Wait — if OA and OC are opposite, then ∠AOC is straight.

So ∠AOB and ∠BOC are adjacent and together make ∠AOC = 180° → so they form a linear pair!

Yes!

So if OA and OC are opposite rays, then:

- ∠AOB and ∠BOC form a linear pair (they are adjacent and together make a straight line)
- Similarly, ∠BOC and ∠COD form a linear pair if OC and OD are opposite? No.

Wait — if OA and OC are opposite, then ∠AOC is straight.

So any angle from OA to OC, passing through OB, will have a linear pair.

So if ray OB is between OA and OC, then:

- ∠AOB and ∠BOC form a linear pair.

Similarly, if ray OD is on the other side, then ∠COD and ∠DOA might be linear.

But let’s assume a common configuration:

Rays: OA, OB, OC, OD, with:
- OA and OC opposite
- OB and OD opposite

Then angles:
- ∠AOB, ∠BOC, ∠COD, ∠DOA

But ∠AOB + ∠BOC = ∠AOC = 180° → linear pair
∠BOC + ∠COD = ∠BOD = 180° → linear pair
etc.

But wait — if OB and OD are opposite, then ∠BOD is straight.

So:

- ∠AOB and ∠BOD? No, not adjacent.

Better: the linear pair for ∠AOB is the angle that shares side OB and has the other side on the opposite ray of OA.

OA’s opposite is OC.

So the angle between OB and OC is ∠BOC.

And ∠AOB and ∠BOC are adjacent and their non-common sides are OA and OC, which are opposite rays.

So yes! They form a linear pair.

Similarly:

- ∠BOC and ∠COD → only if OC and OD are opposite? Not necessarily.

But if OB and OD are opposite, then ∠BOD is straight.

So ∠BOC and ∠COD are adjacent, but their non-common sides are OC and OD — not opposite unless OC and OD are opposite.

So only if the two rays forming the angle are opposite.

So the rule is: two adjacent angles form a linear pair if their non-common sides are opposite rays.

So for ∠AOB:
- Non-common sides: OA and OB
- To be a linear pair, the other angle must share OB and have the other side on the opposite ray of OA → which is OC.

So the other angle is ∠BOC.

So ∠AOB and ∠BOC form a linear pair.

Similarly, ∠BOC and ∠COD → only if OC and OD are opposite? Not necessarily.

But if the diagram has only two lines: say, line AOC and line BOD intersecting at O.

Then:
- Angles: ∠AOB, ∠BOC, ∠COD, ∠DOA

Then:
- ∠AOB and ∠BOC → adjacent, non-common sides OA and OC → opposite rays → linear pair ✔
- ∠BOC and ∠COD → non-common sides OB and OD → opposite rays → linear pair ✔
- ∠COD and ∠DOA → non-common sides OC and OA → opposite → linear pair ✔
- ∠DOA and ∠AOB → non-common sides OD and OB → opposite → linear pair ✔

So every two adjacent angles form a linear pair.

But that’s only if all rays are on two lines.

Now, back to Part B.

Let’s assume the diagram has three lines through O, so six rays.

But for simplicity, let’s assume the common case.

Assumed Configuration for Part B:

Three lines through O:
- Line 1: OA and OD (opposite)
- Line 2: OB and OE (opposite)
- Line 3: OC and OF (opposite)

Then the six angles are:
- ∠AOB, ∠BOC, ∠COD, ∠DOE, ∠EOF, ∠FOA

Now, for each given angle, find the angle that forms a linear pair.

#### 1) ∠AOB

- Adjacent angles: ∠BOC and ∠FOA
- Which one forms a straight line with it?

The non-common sides are OA and OB.
To form a linear pair, the other angle must share one side and have the other side on the opposite ray.

So:
- If we take ∠BOC: shares OB, other side OC → not opposite to OA
- If we take ∠FOA: shares OA, other side OF → not opposite to OB

But OA’s opposite is OD, OB’s opposite is OE.

So the angle that shares side OA and has the other side on the opposite ray of OB (OE) would be ∠AOE.

But ∠AOE is not a single angle; it’s ∠AOB + ∠BOE.

But if there is a ray between B and E, it’s not a single angle.

So unless the figure has only two lines, it’s hard.

Wait — perhaps the figure is simpler: two lines intersecting, forming four angles.

Let’s assume that.

Final Assumption: Two lines intersect at O, forming four angles.

Label the rays: OA, OB, OC, OD, with OA and OC opposite, OB and OD opposite.

Then angles:
- ∠AOB, ∠BOC, ∠COD, ∠DOA

Then:

- ∠AOB and ∠BOC → linear pair (non-common sides OA and OC, opposite rays)
- ∠BOC and ∠COD → linear pair
- ∠COD and ∠DOA → linear pair
- ∠DOA and ∠AOB → linear pair

But for Part B, the questions are:

1) ∠AOB → linear pair with ∠BOC or ∠DOA?

Wait — both are adjacent.

But only one is on the same straight line.

Actually, ∠AOB and ∠BOC are on the same straight line AOC.

So they form a linear pair.

Similarly, ∠AOB and ∠DOA are on the same straight line BOD.

So ∠AOB has two linear pairs: ∠BOC and ∠DOA.

Yes! Because:

- ∠AOB and ∠BOC share side OB, non-common sides OA and OC → opposite rays → linear pair
- ∠AOB and ∠DOA share side OA, non-common sides OB and OD → opposite rays → linear pair

So ∠AOB has two linear pairs: ∠BOC and ∠DOA

Similarly for others.

So let’s answer Part B accordingly.

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Answer Key (Based on Common Worksheet Design)

#### Part A: Identify all linear pairs in each figure

Assume each figure shows intersecting lines.

1) Figure 1: Two lines intersecting at O → four linear pairs:
- ∠AOB and ∠BOC
- ∠BOC and ∠COD
- ∠COD and ∠DOA
- ∠DOA and ∠AOB

(But only if labeled that way)

Better: list all adjacent pairs that are supplementary.

Typically:
- ∠1 and ∠2
- ∠2 and ∠3
- ∠3 and ∠4
- ∠4 and ∠1

2) Figure 2: Similar

But without image, skip.

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#### Part B: Identify the angles that make a linear pair with each specified angle

Assume two lines intersect at O, forming angles:
- ∠AOB, ∠BOC, ∠COD, ∠DOA

With:
- OA and OC opposite
- OB and OD opposite

Then:

1) ∠AOB → linear pairs with ∠BOC (on line AOC) and ∠DOA (on line BOD)
- So: ∠BOC or ∠DOA

2) ∠BOC → linear pairs with ∠AOB (on AOC) and ∠COD (on BOD)
- So: ∠AOB or ∠COD

3) ∠DOC → same as ∠COD
- Linear pairs: ∠BOC (on BOD) and ∠DOA (on AOC)
- So: ∠BOC or ∠DOA

4) ∠AOF → wait, what is F?

Possibly typo or different labeling.

Assume it’s ∠DOA or ∠AOD.

But in the question: ∠AOF

Perhaps ray OF is there.

Maybe the rays are A, B, C, D, E, F.

But to match common worksheets, likely:

- ∠AOF is the angle between OA and OF

But without image, best guess.

Alternatively, assume:

- ∠AOB, ∠BOC, ∠COD, ∠DOE, ∠EOF, ∠FOA

And lines: A-D, B-E, C-F

Then:

1) ∠AOB → linear pair with ∠BOC (if A-B-C on a line) or with ∠FOA (if A-F-O-B on a line)

But only if the non-common sides are opposite.

So for ∠AOB:
- Shares OA: opposite is OD
- Shares OB: opposite is OE
- So the angle that shares OA and has other side on OE? No.

The only way is if the adjacent angle has the other side on the opposite ray.

So for ∠AOB:
- Adjacent angle sharing OA: ∠FOA
- But ∠FOA has sides OF and OA
- Non-common sides: OB and OF — not opposite unless B and F are opposite.

But if B and F are not opposite, not a linear pair.

So only if the two rays are on a straight line.

Final conclusion: in most such worksheets, the answer is:

Standard Answers for Part B:

1) ∠AOB → ∠BOC or ∠DOA
2) ∠BOC → ∠AOB or ∠COD
3) ∠DOC → ∠BOC or ∠DOA
4) ∠AOF → likely typo; perhaps ∠AOB or ∠FOA

But if ∠AOF is between OA and OF, and OF is opposite to OB, then ∠AOF and ∠FOB might be linear, but not listed.

Perhaps the last one is ∠AOD or something.

Given the lack of image, here is a plausible answer set:

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✔ Final Answer (Based on Typical Worksheet):

#### Part A:
1) Linear pairs: ∠AOB & ∠BOC, ∠BOC & ∠COD, ∠COD & ∠DOA, ∠DOA & ∠AOB
2) Similar
3) Similar
4) Similar

#### Part B:
1) ∠AOB → ∠BOC or ∠DOA
2) ∠BOC → ∠AOB or ∠COD
3) ∠DOC → ∠BOC or ∠DOA
4) ∠AOF → ∠FOB or ∠AOD (assuming F and B opposite, A and D opposite)

But since the image isn't visible, I recommend checking the diagram for ray labels.

If you can describe the figure or provide the labels, I can give exact answers.

Otherwise, the above is a general solution for common "Linear Pairs" worksheets.