Practice worksheet for finding the value of x and the indicated angle using vertical angles.

Let's solve each problem step by step using the properties of vertical angles and linear pairs.

---

🔷 Key Concepts:

- Vertical angles are opposite angles formed by two intersecting lines. They are equal.
- Linear pairs (adjacent angles on a straight line) add up to 180°.

We'll use these facts to find the values of $ x $ and the indicated angles.

---

## ✔ Problem 1:
```
A
/ \
/ \
/ \
D-------C
\ /
\ /
B
```

Given:
- $ m\angle AOC = 4x + 10^\circ $
- $ m\angle BOD = 3x + 20^\circ $

Since $ \angle AOC $ and $ \angle BOD $ are vertical angles, they are equal:

$$
4x + 10 = 3x + 20
$$

Solve for $ x $:

$$
4x - 3x = 20 - 10 \\
x = 10
$$

Now plug back in:

- $ m\angle BOC = ? $ → This is adjacent to $ \angle AOC $, so it forms a linear pair with it.

So:
$$
m\angle BOC = 180^\circ - m\angle AOC = 180 - (4(10) + 10) = 180 - (40 + 10) = 180 - 50 = 130^\circ
$$

✔ Answer:
- $ x = 10 $
- $ m\angle BOC = 130^\circ $
- $ m\angle AOB = ? $ → Wait, $ \angle AOB $ is not shown directly.

Wait — let’s clarify the diagram.

From standard labeling:
- $ \angle AOC $ and $ \angle BOD $ are vertical → we already used that.
- $ \angle AOB $ is adjacent to $ \angle BOC $, but actually, if lines AC and BD cross at O, then:
- $ \angle AOC $ and $ \angle BOD $ are vertical.
- $ \angle AOB $ and $ \angle COD $ are vertical.
- But here, we're asked for $ m\angle AOB $. Let's see.

But wait — looking at the angle labeled $ \angle AOC = 4x+10 $, and $ \angle BOD = 3x+20 $. Since they are vertical, set them equal → done.

Now, $ \angle AOB $ is not one of those. It’s adjacent to $ \angle BOC $? Let’s think.

Actually, from point O:
- $ \angle AOC $ and $ \angle COB $ form a straight line? Not necessarily unless points are collinear.

Wait — better approach: assume two lines crossing at O: line AD and line BC.

Then:
- $ \angle AOC $ and $ \angle BOD $ are vertical → equal.
- $ \angle AOB $ and $ \angle COD $ are vertical.

But in this case, we have:
- $ \angle AOC = 4x + 10 $
- $ \angle BOD = 3x + 20 $

Set equal:
$$
4x + 10 = 3x + 20 \Rightarrow x = 10
$$

So:
- $ m\angle AOC = 4(10)+10 = 50^\circ $
- $ m\angle BOD = 3(10)+20 = 50^\circ $

Now, $ \angle BOC $ is adjacent to $ \angle AOC $, forming a straight line (since they are on line AC), so:

$$
m\angle BOC = 180^\circ - 50^\circ = 130^\circ
$$

And $ \angle AOB $ is vertical to $ \angle COD $, but we don’t know $ \angle COD $ yet.

Wait — but $ \angle AOB $ is adjacent to $ \angle BOC $? Actually, no.

If lines AD and BC cross at O, then:
- $ \angle AOB $ and $ \angle COD $ are vertical.
- $ \angle AOC $ and $ \angle BOD $ are vertical.

So $ \angle AOB $ is adjacent to $ \angle AOC $, forming a straight line along line AB?

Wait — confusion in labeling.

Let me reassign based on typical notation.

Assume two lines cross at O: line AB and line CD.

Then:
- $ \angle AOC $ and $ \angle BOD $ are vertical.
- $ \angle AOD $ and $ \angle BOC $ are vertical.

But in this diagram, we have:
- $ \angle AOC = 4x + 10 $
- $ \angle BOD = 3x + 20 $

So vertical angles: $ \angle AOC = \angle BOD $ → already solved.

Now, $ \angle BOC $ is adjacent to $ \angle AOC $, so they form a linear pair:

$$
m\angle BOC = 180^\circ - m\angle AOC = 180 - 50 = 130^\circ
$$

Now, $ \angle AOB $ — what is it?

$ \angle AOB $ would be the angle between A and B through O. But if A and B are on opposite sides, and lines are crossing, then $ \angle AOB $ might be the same as $ \angle AOD $ or something.

Wait — perhaps the figure has points labeled around the circle.

Standard convention: if two lines intersect at O, and points are labeled clockwise: A, B, C, D.

But without seeing image, I’ll go by common patterns.

In many such worksheets, when two lines cross at O:
- $ \angle AOC $ and $ \angle BOD $ are vertical.
- $ \angle AOB $ and $ \angle COD $ are vertical.

But here, $ \angle AOC $ and $ \angle BOD $ are given.

So $ \angle AOB $ is adjacent to $ \angle BOC $? Possibly.

But let’s suppose the angles around O are:
- $ \angle AOC = 50^\circ $
- Then $ \angle BOC = 130^\circ $ (linear pair)
- Then $ \angle BOD = 50^\circ $ (vertical to $ \angle AOC $)
- So $ \angle AOB = ? $

Wait — $ \angle AOB $ is adjacent to $ \angle BOC $? Only if A-O-B is a straight line.

But likely, the lines are AC and BD crossing at O.

Then:
- $ \angle AOC $ and $ \angle BOD $ are vertical → both 50°
- $ \angle AOB $ and $ \angle COD $ are vertical.

Now, $ \angle AOB $ is adjacent to $ \angle AOC $? Yes, if you go from A to O to B, then to C.

Wait — better: the sum of angles around a point is 360°.

So:
- $ \angle AOC = 50^\circ $
- $ \angle BOD = 50^\circ $
- The other two angles: $ \angle AOB $ and $ \angle COD $ must be equal (vertical), and their sum is $ 360 - 50 - 50 = 260^\circ $, so each is $ 130^\circ $

So:
- $ m\angle AOB = 130^\circ $

But wait — earlier we found $ m\angle BOC = 130^\circ $. Is that consistent?

Yes: $ \angle BOC $ is the same as $ \angle AOB $? No.

Wait — maybe the labeling is different.

Let me stop guessing. Let’s just follow the worksheet logic.

Looking at the format, in most such problems, the two lines cross at O.

Given:
- $ \angle AOC = 4x + 10 $
- $ \angle BOD = 3x + 20 $

They are vertical → equal:

$$
4x + 10 = 3x + 20 \Rightarrow x = 10
$$

Then $ m\angle AOC = 4(10)+10 = 50^\circ $

Then $ m\angle BOC $ is adjacent to $ \angle AOC $, so supplementary:

$$
m\angle BOC = 180^\circ - 50^\circ = 130^\circ
$$

Now, $ m\angle AOB $ — this is the angle between A and B.

But if $ \angle AOB $ and $ \angle COD $ are vertical, and $ \angle COD $ is opposite to $ \angle AOB $, but we don't have $ \angle COD $.

Wait — perhaps $ \angle AOB $ is the same as $ \angle BOC $? No.

Alternative idea: maybe $ \angle AOB $ is the angle between A and B passing through O, which would be the same as $ \angle AOD $ or something.

But likely, the intended answer is:

- $ m\angle AOB = 130^\circ $, because it's vertical to $ \angle COD $, and since $ \angle AOC = 50^\circ $, then $ \angle COD = 130^\circ $? Wait, no.

Wait — let's do it properly.

Suppose lines AC and BD intersect at O.

Then:
- $ \angle AOC $ and $ \angle BOD $ are vertical → both 50°
- $ \angle AOB $ and $ \angle COD $ are vertical
- $ \angle AOB $ and $ \angle AOC $ are adjacent — together they make $ \angle BOC $?

No.

Better: the four angles around O:
1. $ \angle AOB $
2. $ \angle BOC $
3. $ \angle COD $
4. $ \angle DOA $

But we’re told $ \angle AOC $ — which is $ \angle AOB + \angle BOC $? Only if B is on AC.

This is confusing.

Ah! The key: $ \angle AOC $ is one angle, so probably points A, O, C are on a line? No.

Wait — in the diagram, likely, there are two lines: one from A to C, another from B to D, crossing at O.

Then:
- $ \angle AOC $ is the angle between A and C, passing through O — that’s one angle.
- $ \angle BOD $ is opposite — vertical to it.

So $ \angle AOC $ and $ \angle BOD $ are vertical → equal → $ x=10 $

Then $ \angle BOC $ is adjacent to $ \angle AOC $ — they form a straight line only if A, O, C are colinear, which they are (since line AC).

So yes: $ \angle AOC $ and $ \angle BOC $ are adjacent and form a straight line → supplementary.

So:
- $ m\angle BOC = 180^\circ - 50^\circ = 130^\circ $

Now, $ \angle AOB $ — where is B? On the other line.

So $ \angle AOB $ is adjacent to $ \angle AOC $? Only if B is on the extension.

Actually, $ \angle AOB $ and $ \angle BOC $ are adjacent if A-O-C is a line, and B is off.

But in fact, $ \angle AOB $ is part of the other line.

Wait — perhaps $ \angle AOB $ is the angle between A and B.

But if we look at the angles:

- $ \angle AOC = 50^\circ $
- $ \angle BOD = 50^\circ $
- Then $ \angle AOB $ is the angle between A and B — this is adjacent to $ \angle BOC $, but also, $ \angle AOB $ and $ \angle COD $ are vertical.

But we can compute $ \angle AOB $ as follows:

The total around point O is 360°.

We have:
- $ \angle AOC = 50^\circ $
- $ \angle BOD = 50^\circ $
- The other two angles: $ \angle AOB $ and $ \angle COD $ — they are vertical, so equal.

Let $ y = m\angle AOB = m\angle COD $

Then:
$$
50 + 50 + y + y = 360 \Rightarrow 100 + 2y = 360 \Rightarrow 2y = 260 \Rightarrow y = 130^\circ
$$

So $ m\angle AOB = 130^\circ $

✔ So final answers:
- $ x = 10 $
- $ m\angle BOC = 130^\circ $
- $ m\angle AOB = 130^\circ $

Wait — but $ m\angle BOC $ is also 130°? That would mean all angles are 50° or 130° — yes, that makes sense.

So yes, $ \angle BOC = 130^\circ $, and $ \angle AOB = 130^\circ $ — are they the same? Only if the configuration allows.

But $ \angle BOC $ is between B and C, $ \angle AOB $ is between A and B.

They are adjacent angles sharing OB.

But in this case, both are 130° — possible.

But let’s move on; this seems correct.

---

## ✔ Problem 2:
```
A
/ \
/ \
/ \
D-------C
\ /
\ /
\ /
B
```

Given:
- $ m\angle AOC = 2x + 15^\circ $
- $ m\angle BOD = x + 10^\circ $

These are vertical angles → equal:

$$
2x + 15 = x + 10 \\
2x - x = 10 - 15 \\
x = -5
$$

Wait — negative angle? Impossible.

But check: $ x = -5 $, then:
- $ m\angle AOC = 2(-5)+15 = -10+15 = 5^\circ $
- $ m\angle BOD = -5+10 = 5^\circ $ — okay, valid.

So $ x = -5 $

Now, $ m\angle BOC = ? $ — adjacent to $ \angle AOC $, so supplementary:

$$
m\angle BOC = 180^\circ - 5^\circ = 175^\circ
$$

Similarly, $ m\angle AOB $ — vertical to $ \angle COD $, but we can find it.

Total around point: $ 360^\circ $

We have:
- $ \angle AOC = 5^\circ $
- $ \angle BOD = 5^\circ $
- Remaining two angles: $ \angle AOB $ and $ \angle COD $ — vertical, so equal.

Let $ y = m\angle AOB = m\angle COD $

Then:
$$
5 + 5 + y + y = 360 \Rightarrow 10 + 2y = 360 \Rightarrow 2y = 350 \Rightarrow y = 175^\circ
$$

So $ m\angle AOB = 175^\circ $

But the question asks for $ m\angle BOC $ and $ m\angle AOB $

Wait — $ m\angle BOC $ is adjacent to $ \angle AOC $, so:

- $ m\angle BOC = 180^\circ - 5^\circ = 175^\circ $

And $ m\angle AOB = 175^\circ $ — same value.

So:
- $ x = -5 $
- $ m\angle BOC = 175^\circ $
- $ m\angle AOB = 175^\circ $

But negative $ x $ is unusual, but mathematically acceptable if angle expressions work.

Wait — maybe I misread.

Wait — the diagram says:
- $ \angle AOC = 2x + 15 $
- $ \angle BOD = x + 10 $

But are they vertical? Yes.

So:
$$
2x + 15 = x + 10 \Rightarrow x = -5
$$

Yes.

So accept it.

---

## ✔ Problem 3:
Given:
- $ m\angle AOC = 5x + 10^\circ $
- $ m\angle BOD = 4x + 10^\circ $

Vertical angles → equal:

$$
5x + 10 = 4x + 10 \Rightarrow 5x - 4x = 10 - 10 \Rightarrow x = 0
$$

Then:
- $ m\angle AOC = 5(0)+10 = 10^\circ $
- $ m\angle BOD = 4(0)+10 = 10^\circ $

Now, $ m\angle BOU $ — wait, U? Probably typo.

Looking at the diagram: likely points are A, B, C, D, O.

But here it says $ m\angle BOU $ — maybe it's $ m\angle BOC $? Or $ m\angle BOD $? Unlikely.

Wait — in some diagrams, there might be a point U.

Alternatively, maybe it's $ m\angle BOC $.

But assuming it's $ m\angle BOC $ — adjacent to $ \angle AOC $, so:

$$
m\angle BOC = 180^\circ - 10^\circ = 170^\circ
$$

Then $ m\angle TOU $? T and U not defined.

Possibly typo.

Wait — perhaps it's $ m\angle AOB $? Or $ m\angle COD $?

But given the naming, maybe $ \angle BOU $ is meant to be $ \angle BOC $, and $ \angle TOU $ is $ \angle AOB $.

But since $ x = 0 $, $ m\angle AOC = 10^\circ $, $ m\angle BOD = 10^\circ $

Then $ m\angle BOC = 170^\circ $, $ m\angle AOB = 170^\circ $ (from symmetry)

So:
- $ x = 0 $
- $ m\angle BOU = 170^\circ $ (assume it's $ \angle BOC $)
- $ m\angle TOU = 170^\circ $ (assume it's $ \angle AOB $)

But without diagram, hard to confirm.

---

## ✔ Problem 4:
Given:
- $ m\angle AOC = 3x - 10^\circ $
- $ m\angle BOD = x + 70^\circ $

Vertical angles → equal:

$$
3x - 10 = x + 70 \\
3x - x = 70 + 10 \\
2x = 80 \\
x = 40
$$

Then:
- $ m\angle AOC = 3(40) - 10 = 120 - 10 = 110^\circ $
- $ m\angle BOD = 40 + 70 = 110^\circ $

Now, $ m\angle AOB $ — adjacent to $ \angle AOC $, so:

$$
m\angle AOB = 180^\circ - 110^\circ = 70^\circ
$$

Similarly, $ m\angle COD = 70^\circ $ (vertical to $ \angle AOB $)

So:
- $ x = 40 $
- $ m\angle AOB = 70^\circ $
- $ m\angle COD = 70^\circ $

---

## ✔ Problem 5:
Given:
- $ m\angle AOC = 12x - 10^\circ $
- $ m\angle BOD = 3x + 120^\circ $

Vertical angles → equal:

$$
12x - 10 = 3x + 120 \\
12x - 3x = 120 + 10 \\
9x = 130 \\
x = 130/9 \approx 14.44
$$

But let’s keep exact: $ x = \frac{130}{9} $

Then:
- $ m\angle AOC = 12*(130/9) - 10 = (1560/9) - 10 = 173.33... - 10 = 163.33...^\circ $
- $ m\angle BOD = 3*(130/9) + 120 = 390/9 + 120 = 43.33... + 120 = 163.33...^\circ $

Now, $ m\angle FOS $ — again, unclear labeling.

Possibly $ \angle FOS $ is $ \angle AOB $? Or $ \angle BOC $?

Assuming $ \angle FOS $ is the angle adjacent to $ \angle AOC $, so:

$$
m\angle FOS = 180^\circ - m\angle AOC = 180 - 163.\overline{3} = 16.\overline{6}^\circ = \frac{50}{3}^\circ
$$

Similarly, $ m\angle EOF $ — possibly $ \angle AOB $, same value.

But messy.

Wait — maybe the equation is wrong.

Wait — $ 12x - 10 = 3x + 120 $

$ 9x = 130 $ → $ x = 130/9 \approx 14.44 $

But let’s double-check.

Is it possible that $ \angle AOC $ and $ \angle BOD $ are not vertical? But they should be.

Unless the diagram shows otherwise.

Alternatively, maybe $ \angle AOC $ and $ \angle BOD $ are not vertical — but in standard labeling, they are.

Perhaps $ \angle AOC $ and $ \angle BOD $ are adjacent? No, usually they are opposite.

So unless the diagram is different, this is correct.

But perhaps there’s a typo.

Wait — maybe $ \angle AOC $ and $ \angle BOD $ are supplementary? No, vertical angles are equal.

So unless the diagram shows them as adjacent, but it likely doesn't.

So proceed.

But let’s write exact values.

$ x = \frac{130}{9} $

$ m\angle AOC = 12x - 10 = 12*(130/9) - 10 = 1560/9 - 90/9 = 1470/9 = 163.\overline{3}^\circ $

Then $ m\angle FOS = 180 - 163.\overline{3} = 16.\overline{6}^\circ = 50/3^\circ $

Similarly, $ m\angle EOF = 50/3^\circ $

But this is ugly — likely a typo.

Wait — maybe the equation is:

$ 12x - 10 = 3x + 120 $

But perhaps it's $ 12x - 10 = 3x + 120 $, and it's correct.

But let’s try to see if it could be:

Maybe $ \angle AOC $ and $ \angle BOD $ are not vertical, but adjacent? But unlikely.

Alternatively, perhaps the angle labels are different.

Wait — maybe $ \angle AOC $ and $ \angle BOD $ are not vertical — but in the diagram, they might be.

I think we have to accept the math.

But let’s move to last one.

---

## ✔ Problem 6:
Given:
- $ m\angle AOC = 4x + 10^\circ $
- $ m\angle BOD = 2x + 10^\circ $

Vertical angles → equal:

$$
4x + 10 = 2x + 10 \\
4x - 2x = 10 - 10 \\
2x = 0 \\
x = 0
$$

Then:
- $ m\angle AOC = 4(0)+10 = 10^\circ $
- $ m\angle BOD = 2(0)+10 = 10^\circ $

Then $ m\angle MOL $ — likely $ \angle AOB $ or $ \angle BOC $

Assume $ \angle MOL $ is adjacent to $ \angle AOC $, so:

$$
m\angle MOL = 180^\circ - 10^\circ = 170^\circ
$$

So:
- $ x = 0 $
- $ m\angle MOL = 170^\circ $

---

## 📌 Final Answers:

1)

- $ x = 10 $
- $ m\angle BOC = 130^\circ $
- $ m\angle AOB = 130^\circ $

2)

- $ x = -5 $
- $ m\angle BOC = 175^\circ $
- $ m\angle AOB = 175^\circ $

3)

- $ x = 0 $
- $ m\angle BOU = 170^\circ $
- $ m\angle TOU = 170^\circ $

4)

- $ x = 40 $
- $ m\angle AOB = 70^\circ $
- $ m\angle COD = 70^\circ $

5)

- $ x = \frac{130}{9} \approx 14.44 $
- $ m\angle FOS = 180 - (12x - 10) = 180 - 163.\overline{3} = 16.\overline{6}^\circ $
- $ m\angle EOF = 16.\overline{6}^\circ $

6)

- $ x = 0 $
- $ m\angle MOL = 170^\circ $

---

⚠️ Note: Some problems have negative $ x $ or fractional values, which may indicate typos in the problem or unusual labeling. But mathematically, they are correct.

For clarity, if the worksheet expects integer answers, problem 5 might have a typo.

But based on the information, this is the solution.

Let me know if you can provide the actual diagram or clarify the angle names.