Arcs, Central Angles & Inscribed Angles Worksheet - Free Printable

Since I can't view or access uploaded images directly, I can’t see the specific diagram or details of your image. However, based on the text you've provided, it appears to be a geometry worksheet focused on arcs, central angles, and inscribed angles in circles.

Let me help you solve this step-by-step using the information given in your text:

---

Given:

- $ m\angle CPX = 28^\circ $
- $ m\angle CPY = 38^\circ $
- $ WZ $ and $ XV $ are diameters.
- $ O $ is the center of the circle in all figures.

We are asked to find various arc measures and angle measures, particularly related to arcs and angles in circles.

Let’s go through each problem systematically.

---

Key Concepts (Review):

1. Central Angle: An angle whose vertex is at the center of the circle. The measure of a central angle is equal to the measure of its intercepted arc.

> $ \text{Measure of arc} = \text{Measure of central angle} $

2. Inscribed Angle: An angle whose vertex is on the circle and whose sides intersect the circle. The measure of an inscribed angle is half the measure of its intercepted arc.

> $ \text{Inscribed angle} = \frac{1}{2} \times \text{intercepted arc} $

3. Diameter: A chord passing through the center. It divides the circle into two semicircles (each 180°).

4. Vertical Angles: Equal in measure.

5. Straight Angle: 180°

6. Arc Addition Postulate: The measure of an arc formed by two adjacent arcs is the sum of the measures of the individual arcs.

---

Now let's solve the problems one by one based on the description.

---

Problem 1: $ \overline{YZ} $

We need to find the measure of arc $ YZ $. Let's analyze what we know.

From the diagram (assumed from standard layout):
- $ \angle CPX = 28^\circ $ → This is a central angle intercepting arc $ CX $, so:
- $ m\overset{\frown}{CX} = 28^\circ $
- $ \angle CPY = 38^\circ $ → Central angle intercepting arc $ CY $, so:
- $ m\overset{\frown}{CY} = 38^\circ $

Now, since $ XV $ and $ WZ $ are diameters, they pass through the center $ O $, so:
- $ XV $ is a diameter → $ \overset{\frown}{XV} = 180^\circ $
- $ WZ $ is a diameter → $ \overset{\frown}{WZ} = 180^\circ $

But more importantly, points $ X, P, Y, Z $ seem to be on the circle with center $ O $. Let’s assume the order around the circle is something like: $ X, C, Y, Z, W, V $, etc., but without the image, we must infer.

Wait — $ \angle CPX $ and $ \angle CPY $ share point $ P $, which might not be the center. But wait — the problem says "O is the center", and the angles are labeled $ \angle CPX $, $ \angle CPY $. So likely, point $ P $ is not the center.

Wait — this is critical. If $ \angle CPX = 28^\circ $, and $ C $, $ P $, $ X $ are points on the circle, then unless $ P $ is the center, this is not a central angle.

But the notation $ \angle CPX $ means vertex at $ P $. So if $ P $ is on the circle, this would be an inscribed angle, not central.

But the problem says $ m\angle CPX = 28^\circ $, and $ m\angle CPY = 38^\circ $. If $ P $ is on the circle, then these are inscribed angles.

But earlier it says “$ WZ $ and $ XV $ are diameters”, and $ O $ is the center.

So likely, point $ P $ is on the circle, and $ C $, $ X $, $ Y $, $ Z $ are also on the circle.

Let’s try to interpret.

Assume that $ C $, $ P $, $ X $, $ Y $, $ Z $, $ W $, $ V $ are points on the circle, and $ O $ is the center.

Given:
- $ m\angle CPX = 28^\circ $ → inscribed angle at $ P $, intercepting arc $ CX $
- So, $ m\angle CPX = \frac{1}{2} m\overset{\frown}{CX} $
→ $ 28^\circ = \frac{1}{2} m\overset{\frown}{CX} $
→ $ m\overset{\frown}{CX} = 56^\circ $

Similarly,
- $ m\angle CPY = 38^\circ $ → inscribed angle at $ P $, intercepting arc $ CY $
→ $ 38^\circ = \frac{1}{2} m\overset{\frown}{CY} $
→ $ m\overset{\frown}{CY} = 76^\circ $

Now, since both arcs $ CX $ and $ CY $ start at $ C $, and go to $ X $ and $ Y $, and assuming $ X $ and $ Y $ are on the same side of $ C $, we can find arc $ XY $ as the difference?

But wait — if $ \angle CPX $ intercepts arc $ CX $, and $ \angle CPY $ intercepts arc $ CY $, and both have vertex at $ P $, then $ P $ must be on the opposite side of the circle from arc $ CX $ and $ CY $.

Standard rule: an inscribed angle intercepts the arc opposite to it.

So, for $ \angle CPX $, the intercepted arc is $ \overset{\frown}{CX} $, meaning arc $ CX $ not containing $ P $.

Similarly, $ \angle CPY $ intercepts arc $ \overset{\frown}{CY} $.

So, $ m\overset{\frown}{CX} = 2 \times 28^\circ = 56^\circ $

$ m\overset{\frown}{CY} = 2 \times 38^\circ = 76^\circ $

Now, if $ X $, $ C $, $ Y $ are in order around the circle, then arc $ XY $ = arc $ CY $ − arc $ CX $ = 76° − 56° = 20°? Only if $ X $ is between $ C $ and $ Y $.

But we don’t know the order.

Alternatively, maybe $ C $ is common, and $ X $ and $ Y $ are on different sides.

But perhaps there’s another interpretation.

Wait — maybe $ P $ is not on the circle? But the angle is named $ \angle CPX $, with vertex at $ P $, and if $ P $ were the center, it would be a central angle.

But the problem says $ O $ is the center, so $ P $ is probably not the center.

Unless $ P $ is $ O $? But $ O $ is the center, and $ P $ is used in angle names.

Possibility: Maybe $ P $ is a typo, or $ P $ is $ O $? But the problem says “O is the center”, so likely $ P $ is a point on the circle.

But then $ \angle CPX $ is an inscribed angle.

But let’s suppose instead that $ \angle CPX $ and $ \angle CPY $ are central angles. That would make sense only if $ P $ is the center.

But the problem says “O is the center”, so $ P $ cannot be the center unless $ P = O $.

So perhaps $ P $ is $ O $? But the labels are different.

This is ambiguous.

Wait — look at the figure references: in problem 17, 18, etc., there is a point $ O $, and $ P $ is sometimes used.

But in the first part, it says:

> Find each measure in $ \odot P $ if $ m\angle CPX = 28^\circ $, $ m\angle CPY = 38^\circ $, and $ WZ $, $ XV $ are diameters.

Ah! Here’s the key: $ \odot P $ means the circle with center $ P $.

So $ P $ is the center of the circle!

That changes everything.

So $ P $ is the center, and $ O $ is mentioned later — but maybe $ O $ is just another name, or perhaps it's a typo.

Wait — no: it says “in each of the following figures, $ O $ is the center” — so possibly in the first part, $ P $ is the center, but in the rest, $ O $ is the center.

But in the first part, it says “in $ \odot P $”, so $ P $ is the center.

So for questions 1–7, the center is $ P $, and for 8–20, the center is $ O $.

So let's proceed.

---

Part 1: Circle with center $ P $

Given:
- $ m\angle CPX = 28^\circ $ → central angle → $ m\overset{\frown}{CX} = 28^\circ $
- $ m\angle CPY = 38^\circ $ → central angle → $ m\overset{\frown}{CY} = 38^\circ $
- $ WZ $ and $ XV $ are diameters → so $ \overset{\frown}{WZ} = 180^\circ $, $ \overset{\frown}{XV} = 180^\circ $

Now solve:

#### 1. $ \overset{\frown}{YZ} $

We need arc $ YZ $. We know:
- $ \overset{\frown}{CX} = 28^\circ $
- $ \overset{\frown}{CY} = 38^\circ $

But $ C $ is a common point.

Assuming points are arranged such that $ X $, $ C $, $ Y $, $ Z $ are in order around the circle.

But we don't know the full configuration.

But we know $ XV $ is a diameter → $ \overset{\frown}{XV} = 180^\circ $

Similarly, $ WZ $ is a diameter → $ \overset{\frown}{WZ} = 180^\circ $

But we need more info.

Wait — since $ \angle CPX = 28^\circ $, and $ P $ is center, then $ \angle CPX $ is the angle at $ P $ between points $ C $, $ P $, $ X $, so it intercepts arc $ CX $, so yes, $ m\overset{\frown}{CX} = 28^\circ $

Similarly, $ m\overset{\frown}{CY} = 38^\circ $

Now, if $ X $, $ C $, $ Y $ are on the circle, and $ P $ is center, then the arc $ XY $ could be $ |38^\circ - 28^\circ| = 10^\circ $, but only if $ C $ is between $ X $ and $ Y $.

But we don’t know.

Alternatively, maybe $ \angle CPX $ and $ \angle CPY $ are adjacent angles at the center.

Suppose rays $ PC $, $ PX $, $ PY $ are drawn.

Then $ \angle CPX = 28^\circ $, $ \angle CPY = 38^\circ $, so the angle between $ PX $ and $ PY $ depends on whether $ Y $ is on the same side as $ X $ relative to $ C $.

But likely, $ \angle XPY = \angle CPY - \angle CPX = 38^\circ - 28^\circ = 10^\circ $, if $ X $ is between $ C $ and $ Y $.

So $ m\overset{\frown}{XY} = 10^\circ $

But we need $ \overset{\frown}{YZ} $

We need more.

Perhaps $ WZ $ is a diameter, so $ W $ and $ Z $ are endpoints of a diameter.

Similarly, $ XV $ is a diameter.

So $ X $ and $ V $ are endpoints of a diameter.

So $ \overset{\frown}{XV} = 180^\circ $

Similarly, $ \overset{\frown}{WZ} = 180^\circ $

Now, if $ \overset{\frown}{CX} = 28^\circ $, and $ \overset{\frown}{CY} = 38^\circ $, and if $ C $ is between $ X $ and $ Y $, then $ \overset{\frown}{XY} = 10^\circ $

But still not helping.

Maybe $ \angle CPX = 28^\circ $, so arc $ CX = 28^\circ $

$ \angle CPY = 38^\circ $, so arc $ CY = 38^\circ $

Now, if $ X $ and $ Y $ are on the same side of $ C $, then arc $ XY = |38 - 28| = 10^\circ $

But we need arc $ YZ $.

Perhaps $ Z $ is on the other end of diameter $ WZ $.

But without diagram, hard.

Alternatively, maybe the total circle is 360°, and we can use symmetry.

But let’s move to problems where we have diagrams.

---

Since I can't see the diagrams for problems 8–20, I can't give exact answers, but I can explain how to solve them.

But let’s try to do problem 8.

---

Problem 8: In a circle with center $ O $, triangle $ ABC $, $ AB $ is diameter, $ \angle ACB = ? $

Wait, no — problem 8 shows a circle with center $ O $, points $ A $, $ B $, $ C $, $ D $, and $ \angle AOB = 100^\circ $, $ \angle COD = ? $, but it's unclear.

Since I can't see the images, I can only provide general solutions.

---

General Approach for Problems 8–20:

Each problem involves finding an arc or angle in a circle with center $ O $.

Use these rules:

1. Central angle = intercepted arc
2. Inscribed angle = half the intercepted arc
3. Angle formed by two chords intersecting inside circle:
$$
\text{Angle} = \frac{1}{2} (\text{arc}_1 + \text{arc}_2)
$$
4. Angle formed by two secants from outside:
$$
\text{Angle} = \frac{1}{2} |\text{arc}_1 - \text{arc}_2|
$$
5. Tangent-chord angle = half the intercepted arc
6. Diameters create 180° arcs

---

Example: Problem 17

Typical problem: Two chords intersect at a point inside the circle. You're given some angles or arcs.

For example, if two chords $ AB $ and $ CD $ intersect at $ P $, then:
$$
\angle APC = \frac{1}{2} (\overset{\frown}{AC} + \overset{\frown}{BD})
$$

If you're given one arc, you may need to find others using the fact that the whole circle is 360°.

---

Example: Problem 19

Often shows a right triangle inscribed in a semicircle (Thales' theorem): if $ AB $ is diameter, and $ C $ is on circle, then $ \angle ACB = 90^\circ $

Or, given $ \angle AOB = 100^\circ $, find $ \angle ACB $ where $ C $ is on the circle.

Then $ \angle ACB = \frac{1}{2} \overset{\frown}{AB} = \frac{1}{2} \times 100^\circ = 50^\circ $

---

Problem 20

Shows two arcs: $ \overset{\frown}{AB} = 120^\circ $, $ \overset{\frown}{CD} = 101^\circ $, and an angle at $ P $, maybe intersection point.

Then the angle at $ P $ is:
$$
\angle APB = \frac{1}{2} (\overset{\frown}{AB} + \overset{\frown}{CD}) = \frac{1}{2}(120 + 101) = \frac{221}{2} = 110.5^\circ
$$

But depends on configuration.

---

Back to First Part: Questions 1–7

Assuming $ P $ is center, $ \angle CPX = 28^\circ $ → arc $ CX = 28^\circ $

$ \angle CPY = 38^\circ $ → arc $ CY = 38^\circ $

And $ XV $, $ WZ $ are diameters.

So $ \overset{\frown}{XV} = 180^\circ $, $ \overset{\frown}{WZ} = 180^\circ $

Now, let’s try to find arc $ YZ $.

But we need to know where $ Y $ and $ Z $ are.

Perhaps $ C $, $ X $, $ Y $, $ Z $ are in order.

But if $ \overset{\frown}{CX} = 28^\circ $, $ \overset{\frown}{CY} = 38^\circ $, then if $ X $ is between $ C $ and $ Y $, then $ \overset{\frown}{XY} = 10^\circ $

But we need $ \overset{\frown}{YZ} $

Without knowing where $ Z $ is, we can't say.

But perhaps $ WZ $ is a diameter, so $ W $ and $ Z $ are endpoints.

Similarly, $ XV $ is a diameter.

So $ X $ and $ V $ are endpoints.

So $ \overset{\frown}{XV} = 180^\circ $

So if $ \overset{\frown}{CX} = 28^\circ $, then $ \overset{\frown}{CV} = 180^\circ - 28^\circ = 152^\circ $, since $ X $ and $ V $ are ends of diameter.

Similarly, if $ \overset{\frown}{CY} = 38^\circ $, then $ \overset{\frown}{YV} = 180^\circ - 38^\circ = 142^\circ $, if $ Y $ is on the same arc.

But still not helpful.

Perhaps $ \angle CPX = 28^\circ $, $ \angle CPY = 38^\circ $, and they are adjacent, so $ \angle XPY = 38^\circ - 28^\circ = 10^\circ $, so arc $ XY = 10^\circ $

But again, not $ YZ $.

Perhaps $ Z $ is such that $ WZ $ is diameter, and $ W $ is opposite $ Z $.

But without diagram, impossible.

---

Conclusion:

I can't solve the specific problems without seeing the diagrams.

However, here’s how to approach each type:

- Find arc measure: Use central angle = arc measure.
- Find inscribed angle: Half the arc.
- Two chords intersecting: Angle = half the sum of intercepted arcs.
- Secants from outside: Angle = half the difference of intercepted arcs.
- Tangent and chord: Angle = half the intercepted arc.
- Diameter: Creates 180° arc.
- Right angle in semicircle: Thales’ theorem.

---

If you can describe the diagrams for problems 1–7 and 8–20, I can solve them exactly.

For example, for problem 1: Is $ \overset{\frown}{YZ} $ the arc from $ Y $ to $ Z $, and is $ Y $ and $ Z $ connected by a diameter? Or is $ C $ between $ Y $ and $ Z $?

Please describe the figure or upload a clear image.

Alternatively, if you can list the questions with descriptions, I can help.

But based on current info, here’s what we know:

From $ \angle CPX = 28^\circ $, $ \angle CPY = 38^\circ $, and $ P $ is center:

- $ m\overset{\frown}{CX} = 28^\circ $
- $ m\overset{\frown}{CY} = 38^\circ $
- So $ m\overset{\frown}{XY} = |38 - 28| = 10^\circ $ (if $ X $ and $ Y $ are on the same side of $ C $)

But without more, can't find $ \overset{\frown}{YZ} $

Similarly, for other problems.

---

Final Answer:

Due to the lack of visual access to the diagrams, I cannot provide exact numerical answers for each problem. However, I can guide you through the solution process using geometric principles.

Please describe the diagrams or provide the following for each problem:

- What points are on the circle?
- Where is the center?
- What arcs or angles are given?
- What is being asked?

With that, I can solve each one.

Alternatively, if you can type out the description of each figure, I’ll solve it step by step.

For now, here’s a sample answer for problem 1:

> Problem 1: $ \overset{\frown}{YZ} $
>
> Given $ m\angle CPX = 28^\circ $, and $ P $ is the center, then $ m\overset{\frown}{CX} = 28^\circ $
>
> Given $ m\angle CPY = 38^\circ $, so $ m\overset{\frown}{CY} = 38^\circ $
>
> Assuming $ X $, $ C $, $ Y $ are in order, then $ m\overset{\frown}{XY} = 38^\circ - 28^\circ = 10^\circ $
>
> But without knowing the position of $ Z $, we cannot determine $ m\overset{\frown}{YZ} $
>
> However, if $ WZ $ is a diameter, and $ C $ is on the circle, and if $ \overset{\frown}{CZ} $ is part of the semicircle, we may need more.
>
> Insufficient information without diagram.

Same for others.

Please provide more details.