Vertical Angles Worksheets - Free Printable

Let’s solve each problem one by one. Remember: vertical angles are always equal. Also, angles on a straight line add up to 180°.

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Problem 1:

We see two lines crossing. The angle labeled “(2x + 55)°” and the angle labeled “165°” are vertical angles? Wait — no! Actually, look carefully: the 165° is opposite the (2x + 55)° angle? Let me check the diagram mentally.

Actually, in Problem 1, the angle marked 165° is vertically opposite to the angle marked (2x + 55)°? No — wait, if they’re vertical, they should be equal. But 2x + 55 = 165? That would make sense.

Wait — actually, looking at standard vertical angle diagrams: when two lines cross, the angles across from each other are vertical and equal.

In Problem 1: the angle labeled (2x + 55)° and the angle labeled 165° are not vertical — they are adjacent? Or maybe they are vertical?

Hold on — let’s think again. If two lines intersect, they form two pairs of vertical angles.

Looking at the labels: points D, C, A, B around the intersection. Angle DOC is given as (2x + 55)°, and angle AOB is 165°. Since AOB and DOC are opposite each other, they ARE vertical angles → so they must be equal.

So:

> 2x + 55 = 165

Subtract 55 from both sides:

> 2x = 110

Divide by 2:

> x = 55

Now, m∠COB — that’s the angle next to DOC. Since DOC and COB are on a straight line (line DB), they add to 180°.

DOC = 165°, so COB = 180 - 165 = 15°

✔ So for #1:
x = 55
m∠COB = 15°

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Problem 2:

Angle UOT is labeled (7x - 9)°, and angle VOS is labeled 40°. Are these vertical? Points: U, T, V, S — assuming lines UT and VS cross at O.

Then angle UOT and angle VOS are vertical? Yes — because they are opposite each other.

So:

> 7x - 9 = 40

Add 9 to both sides:

> 7x = 49

Divide by 7:

> x = 7

Now, m∠UOT = 7x - 9 = 7*7 - 9 = 49 - 9 = 40° (which matches)

But the question asks for m∠UOT — which we already know is 40°, but let’s confirm with x=7.

Yes.

✔ So for #2:
x = 7
m∠UOT = 40°

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Problem 3:

Angle ROX is labeled (8x - 6)°, and angle SOQ is labeled 130°. Are they vertical?

Points R, X, S, Q — lines RX and SQ cross at O.

Angle ROX and angle SOQ — are they opposite? Let’s see: if you go from R to O to X, and S to O to Q — depending on labeling, but typically in such diagrams, ROX and SOQ are vertical angles.

Assume yes → then:

> 8x - 6 = 130

Add 6:

> 8x = 136

Divide by 8:

> x = 17

Now, m∠ROX = 8x - 6 = 8*17 - 6 = 136 - 6 = 130° — correct.

✔ So for #3:
x = 17
m∠ROX = 130°

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Problem 4:

Angle HOG is labeled (13x + 64)°, and angle FOI is labeled 146°. Are they vertical?

Points H, G, F, I — lines HG and FI cross at O.

Angle HOG and angle FOI — if they are opposite, then vertical → equal.

So:

> 13x + 64 = 146

Subtract 64:

> 13x = 82

Wait — 146 - 64 = 82? Let me calculate: 146 - 60 = 86, minus 4 more is 82. Yes.

So:

> 13x = 82

x = 82 / 13 ≈ 6.307...? That doesn’t seem right for a middle school problem. Did I misidentify the angles?

Wait — maybe angle HOG and angle FOI are NOT vertical? Maybe they are adjacent?

Look again: perhaps angle HOG and angle FOI are on a straight line? Or maybe vertical angles are different.

Alternative: maybe angle HOG and angle IOF are vertical? Same thing.

Wait — perhaps the 146° is not vertical to (13x+64). Maybe it's supplementary?

Another possibility: maybe angle HOG and the 146° angle are adjacent and form a straight line? Then they’d add to 180.

Try that:

If (13x + 64) + 146 = 180

Then:

13x + 210 = 180

13x = 180 - 210 = -30 → negative? Not possible.

That can’t be.

Wait — perhaps the 146° is vertical to another angle, and (13x+64) is vertical to something else?

Maybe I need to re-express.

Standard approach: in intersecting lines, vertical angles are equal, and adjacent angles sum to 180.

Suppose angle HOG = (13x + 64)°, and angle FOI = 146° — if they are vertical, set equal.

But 13x + 64 = 146 → 13x = 82 → x = 82/13 ≈ 6.307 — not nice.

Perhaps the 146° is the measure of angle FOH or something else?

Wait — maybe angle HOG and angle FOI are NOT vertical — maybe angle HOG and angle IOF are the same, but perhaps the 146° is angle GOI or something.

This is tricky without seeing the exact diagram, but based on common problems, likely they are vertical.

But 82 divided by 13 is not integer — maybe I made a mistake.

Wait — let me double-check subtraction: 146 - 64.

64 + 82 = 146? 64 + 80 = 144, +2 = 146 — yes.

But 82 ÷ 13 = 6.307... — unlikely for this level.

Alternative idea: perhaps the 146° is the measure of the angle adjacent to HOG, so they are supplementary.

So:

(13x + 64) + 146 = 180

13x + 210 = 180

13x = -30 — impossible.

No.

Perhaps the 146° is vertical to a different angle, and (13x+64) is vertical to another.

Wait — maybe angle HOG and angle FOI are not the vertical pair; maybe angle HOG and angle IOF are the same, but perhaps the 146° is angle HOF or something.

I think there might be a mislabeling in my assumption.

Let me try a different approach: in many textbooks, for problem 4, often the angle given is vertical to the expression.

Perhaps it's 13x + 64 = 146, and we accept fractional answer? But that seems odd.

Wait — let me calculate 13*6 = 78, 78 + 64 = 142 — close to 146.

13*6.3 = 81.9, +64 = 145.9 — approximately 146.

But probably not.

Another thought: maybe the 146° is the measure of angle GOI, and angle HOG is vertical to angle IOF, but that doesn't help.

Perhaps the angle labeled 146° is actually the vertical angle to (13x+64), so we have to go with it.

But let's check problem 5 and 6 first, come back.

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Problem 5:

Angle ROS is labeled (11 - 3x)°, and angle QOP is labeled 110°. Are they vertical?

Points R, S, Q, P — lines RS and QP cross at O.

Angle ROS and angle QOP — if they are opposite, then vertical → equal.

So:

> 11 - 3x = 110

Then:

-3x = 110 - 11 = 99

x = 99 / (-3) = -33

Negative? That can't be right for an angle measure.

Angle can't be negative. So probably not vertical.

Perhaps they are adjacent? On a straight line?

So (11 - 3x) + 110 = 180

Then:

121 - 3x = 180

-3x = 180 - 121 = 59

x = -59/3 — still negative.

Worse.

Perhaps angle ROS and angle QOP are vertical, but the expression is for a different angle.

Another possibility: maybe (11 - 3x) is not the angle itself, but part of it? Unlikely.

Or perhaps it's |11 - 3x|, but no.

Wait — maybe the 110° is not angle QOP, but angle SOP or something.

Let's think differently. In some diagrams, the angle labeled might be the reflex or something, but unlikely.

Perhaps for problem 5, angle ROS and angle QOP are vertical, but the equation is 11 - 3x = 110, which gives negative, so maybe it's 3x - 11 = 110? But the label says (11 - 3x).

Unless it's a typo, but we have to work with what's given.

Another idea: perhaps the angle is (3x - 11)°, but written as (11 - 3x) by mistake? Because otherwise it's negative.

Let me assume that. Suppose it's (3x - 11) = 110.

Then 3x = 121, x = 121/3 ≈ 40.333 — still not nice.

Or perhaps it's supplementary.

Let me try: if angle ROS and angle QOP are adjacent, sum to 180.

So (11 - 3x) + 110 = 180

As before, 121 - 3x = 180, -3x = 59, x = -59/3 — no.

Perhaps angle ROS is vertical to another angle, and 110° is given for a different purpose.

I recall that in some problems, the angle might be expressed as (11 - 3x) but it's meant to be positive, so perhaps x is small.

For example, if x=0, angle=11°, too small.

If x=3, 11-9=2°.

Still small.

Perhaps the 110° is the vertical angle, so 11 - 3x = 110 is wrong; maybe it's the other way.

Another thought: in problem 5, the angle labeled (11 - 3x)° might be angle QOR or something, and 110° is angle SOP, and they are vertical.

But same issue.

Perhaps it's not vertical angles, but linear pair.

Let's look at the points: R, O, S and Q, O, P. If lines are RQ and SP crossing at O, then angle ROS and angle QOP might be vertical.

But mathematically, 11 - 3x = 110 has no good solution.

Unless the 110° is not the measure of the vertical angle, but of an adjacent angle.

Suppose angle ROS and angle ROQ are adjacent, sum to 180, and angle ROQ is 110°, then angle ROS = 70°.

So 11 - 3x = 70

-3x = 59

x = -59/3 — still bad.

If angle ROS = 11 - 3x, and it's vertical to an angle that is 110°, then 11 - 3x = 110, same thing.

I think there might be a typo in the problem, or in my interpretation.

Let me skip and do problem 6.

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Problem 6:

Angle MOL is labeled (8x - 17)°, and angle NOP is labeled 79°. Are they vertical?

Points M, L, N, P — lines ML and NP cross at O.

Angle MOL and angle NOP — if they are opposite, then vertical → equal.

So:

> 8x - 17 = 79

Add 17:

> 8x = 96

Divide by 8:

> x = 12

Then m∠MOL = 8*12 - 17 = 96 - 17 = 79° — correct.

✔ So for #6:
x = 12
m∠MOL = 79°

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Now back to problem 4 and 5.

For problem 4: 13x + 64 = 146

13x = 82

x = 82/13 = 6.307... — but perhaps it's acceptable? Or maybe I have the wrong pairing.

Another possibility: in problem 4, the 146° might be the measure of angle HOF or something, and angle HOG is vertical to angle IOF, but angle IOF might be different.

Perhaps angle HOG and angle FOI are not vertical; maybe angle HOG and angle GOI are adjacent.

Let's assume that angle HOG and the 146° angle are on a straight line, so supplementary.

So (13x + 64) + 146 = 180

13x + 210 = 180

13x = -30 — impossible.

If they are vertical, x = 82/13.

But 82/13 is approximately 6.307, and 13*6.307 = 82, +64 = 146, so it works numerically, even if not integer.

Perhaps in the diagram, it's correct.

For problem 5: 11 - 3x = 110

-3x = 99

x = -33

Then angle ROS = 11 - 3*(-33) = 11 + 99 = 110° — oh! Wait a minute!

If x = -33, then 11 - 3*(-33) = 11 + 99 = 110°, which equals the given 110°.

And since they are vertical angles, they should be equal, so it works!

Even though x is negative, the angle measure is positive, which is fine. Angles can't be negative, but the variable x can be negative as long as the angle expression gives a positive number.

So for problem 5:

11 - 3x = 110

-3x = 99

x = -33

Then m∠ROS = 11 - 3*(-33) = 11 + 99 = 110°

Perfect.

Similarly, for problem 4, if we allow x = 82/13, then angle is 146°, which is fine.

But 82/13 is 6 and 4/13, or approximately 6.307, but perhaps leave as fraction.

82/13 is already simplified.

But let's see if it reduces — 82 and 13, 13 is prime, 82÷13 not integer, so x = 82/13.

But in many textbooks, they avoid fractions, so perhaps I have the wrong pairing for problem 4.

Another idea for problem 4: perhaps the 146° is not angle FOI, but angle HOF or something else.

Maybe angle HOG and angle FOI are not vertical; perhaps angle HOG is vertical to angle IOF, and angle FOI is the same as angle IOF, so same thing.

Perhaps the 146° is the measure of the adjacent angle.

Let's calculate what x should be if it's nice number.

Suppose x=6, then 13*6 +64 = 78+64=142, close to 146.

x=6.3, 13*6.3=81.9, +64=145.9≈146.

So probably x=82/13 is intended.

Perhaps it's 13x + 64 = 146, and we box the fraction.

For consistency, let's proceed.

So for problem 4:

13x + 64 = 146

13x = 82

x = 82/13

m∠HOG = 146° (since vertical to 146°)

The question asks for m∠HOG, which is the same as the vertical angle, so 146°.

Similarly for others.

Now summarize all:

Problem 1:
x = 55
m∠COB = 15°

Problem 2:
x = 7
m∠UOT = 40°

Problem 3:
x = 17
m∠ROX = 130°

Problem 4:
x = 82/13
m∠HOG = 146°

Problem 5:
x = -33
m∠ROS = 110°

Problem 6:
x = 12
m∠MOL = 79°

For problem 4, perhaps write x as mixed number: 82 ÷ 13 = 6 with remainder 4, so 6 4/13.

But usually in math, improper fraction is fine.

Since the other answers are integers, maybe I missed something.

Another thought for problem 4: perhaps the 146° is not the vertical angle, but the angle adjacent, and they sum to 180.

But earlier calculation gave negative.

Unless the expression is for the other angle.

Suppose that angle HOG = (13x + 64)°, and it is adjacent to the 146° angle, so:

(13x + 64) + 146 = 180

13x + 210 = 180

13x = -30 — impossible.

If the 146° is vertical to a different angle, say angle GOI, and angle HOG is vertical to angle IOF, but then we don't have information.

I think we have to go with the vertical angle assumption.

Perhaps in problem 4, the angle labeled 146° is angle FOH, and angle HOG is part of it, but unlikely.

I'll stick with x = 82/13 for problem 4.

But let's double-check problem 5: with x=-33, angle is 110°, which is positive, so ok.

Similarly for problem 4, x=82/13≈6.307, angle=146°, positive.

So it's acceptable.

Now for the final answer, we need to provide for each problem.

The user asked to solve the problem, and there are six problems.

So I'll list them.

Final Answer:

1) x = 55, m∠COB = 15°
2) x = 7, m∠UOT = 40°
3) x = 17, m∠ROX = 130°
4) x = \frac{82}{13}, m∠HOG = 146°
5) x = -33, m∠ROS = 110°
6) x = 12, m∠MOL = 79°

For problem 4, perhaps simplify or write as decimal, but fraction is exact.

Since the instruction is to provide final answer, and it's multiple parts, I'll box each.

But the format says "Final Answer:" and then the answer.

Perhaps list them clearly.

Since it's a worksheet, probably expect the values.

I think it's fine.

So here goes.