Practice worksheet on central and inscribed angles with circle diagrams and problem-solving exercises.

Since I can't view or access images directly, I can’t see the specific diagram or details of the problems in your uploaded image. However, based on the text you've provided, I can help you solve the Central Angles and Inscribed Angles practice problems step by step.

Let’s go through each problem logically, assuming standard circle geometry rules:

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Key Concepts:

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.

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

3. Arc Measure: The measure of an arc is equal to the measure of its central angle.

4. Inscribed Angle Theorem:
$$
\text{Inscribed Angle} = \frac{1}{2} \times \text{Intercepted Arc}
$$

5. Opposite Angles in a Cyclic Quadrilateral:
- Sum of opposite angles = 180° (if applicable).

6. Angle formed by two chords intersecting inside a circle:
$$
\text{Angle} = \frac{1}{2}(\text{arc}_1 + \text{arc}_2)
$$

7. Tangent-Chord Angle:
$$
\text{Angle} = \frac{1}{2} \times \text{intercepted arc}
$$

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Now, let's work through the problems as described.

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Problems 1–3: Refer to the figure with O as center

Assuming there is a circle with center O and points A, B, C on the circumference.

#### 1. Name the chord(s)
A chord is a line segment connecting two points on the circle.

- Examples: AB, BC, AC, etc., depending on the figure.
- Likely answer: AB, BC, AC — any two points connected by a straight line within the circle.

> ✔ Answer: Chords: AB, BC, AC (or whatever segments connect two points on the circle)

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#### 2. Name the central angle(s)
A central angle has its vertex at the center O.

- So if OA and OB are radii forming an angle at O, then ∠AOB is a central angle.

> ✔ Answer: ∠AOB, ∠BOC, ∠AOC — depending on labeled points.

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#### 3. Name the inscribed angle(s)
An inscribed angle has its vertex on the circle.

- Example: ∠ABC, where point B is on the circle, and A and C are also on the circle.

> ✔ Answer: ∠ABC, ∠BAC, ∠ACB — any angle with vertex on the circle.

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Problems 4–7: Find indicated measures

We’ll assume typical configurations for these types of problems.

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#### 4. Given: m∠COB = ?; m∠CAB = ?

Let’s suppose we have a circle with center O, and points A, B, C on the circle.

If arc AB is given (say 60°), then:

- Central angle ∠AOB = arc AB = 60°
- Inscribed angle ∠ACB = ½ × arc AB = 30°

But since no values are given, perhaps the diagram shows something like:

Suppose arc AB = 80°, then:

- m∠AOB = 80° (central angle)
- m∠ACB = ½ × 80° = 40° (inscribed angle)

But without numbers, we need to guess from common setups.

Wait — maybe the diagram shows a diameter or right angles?

Alternatively, let’s suppose:

> Suppose arc AB = 100°, then:
> - m∠AOB = 100°
> - m∠ACB = 50°

But again, without values, this is speculative.

Let’s look at the next ones.

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#### 5. m∠EBD = ?; m∠EFD = ?

This suggests a circle with points E, B, D, F.

Possibly quadrilateral inscribed in a circle.

Assume:

- ∠EBD is an inscribed angle intercepting arc ED
- ∠EFD is another inscribed angle intercepting arc ED

Then both would be half the arc.

But unless arcs are marked, we can’t compute.

Wait — perhaps it's a semicircle? Or right triangle?

Common trick: If a triangle is inscribed in a semicircle, the angle at the circumference is 90°.

So if BD is a diameter, and E is on the circle, then ∠BED = 90°.

But still, without values, we need to infer.

Perhaps the diagram shows:

- Arc ED = 120° → then inscribed angle = 60°
- But again, not sure.

Let me switch to problems 8–9, which ask for unknown values.

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#### 8. m∠CED = ?; m∠CEA = ?

Suppose C, E, D, A are on the circle.

Maybe CE is a diameter?

Or perhaps triangle CED has some known arcs.

Without the figure, here's a common setup:

Suppose arc CD = 100°, then:

- m∠CED (inscribed) = ½ × arc CD = 50°

Similarly, if arc CA = 140°, then ∠CEA = 70°

But again, we’re guessing.

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#### 9. m∠POQ = ?; m∠PQO = ?

Assume O is center, P and Q on circle.

Then ∠POQ is central angle → equals arc PQ.

Suppose arc PQ = 120°, then ∠POQ = 120°

Then triangle POQ: OP = OQ (radii), so isosceles.

Then ∠PQO = (180° − 120°)/2 = 30°

So possible answers:

- m∠POQ = 120°
- m∠PQO = 30°

But depends on diagram.

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Find the unknown value: Problems 10–11

> "The figure shows a passenger airplane’s flight path on a circular radar screen as an arc of a circle centered at O."

So we have a circle with center O, and points M, T, N, K on the circle.

Let’s suppose:

#### 10. What is m∠M?

Assume ∠M is an inscribed angle.

Suppose it intercepts arc KT.

Then m∠M = ½ × arc KT

If arc KT = 100°, then ∠M = 50°

But again, need arc measure.

Wait — perhaps the diagram shows that arc MT = 120°, and ∠M is at point M?

No — if ∠M is at point M, and sides go to K and T, then it intercepts arc KT.

So yes: m∠M = ½ × arc KT

Unless it's a tangent, but likely inscribed.

#### 11. What is m∠KMT?

This is angle at M between points K, M, T.

So ∠KMT is an inscribed angle intercepting arc KT.

So m∠KMT = ½ × arc KT

Same as above.

#### 12. What is m∠NMT?

Point N, M, T — so angle at M between N and T.

So ∠NMT intercepts arc NT.

So m∠NMT = ½ × arc NT

Again, depends on arc measures.

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Common Problem Type: Airplane Radar Screen

Typical setup:

- Circle with center O
- Points M, T, K, N on circle
- Airplane flies from K to T, so arc KT is shown
- ∠KOT is central angle
- ∠KMT is inscribed angle

Example:

Suppose arc KT = 100°

Then:

- m∠KOT = 100° (central angle)
- m∠KMT = ½ × 100° = 50° (inscribed angle)

So:

- Problem 10: m∠M = ? → likely refers to ∠KMT = 50°
- Problem 11: m∠KMT = ? → 50°
- Problem 12: m∠NMT = ? → if arc NT = 60°, then 30°

But without actual values, we can only generalize.

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Final Answer (Based on Common Problems):

Let’s assume a standard configuration where:

- Arc KT = 100°
- Arc NT = 60°
- O is center

Then:

#### 10. m∠M = ?
→ Assuming ∠M is ∠KMT, inscribed angle intercepting arc KT
→ m∠KMT = ½ × 100° = 50°

✔ Answer: 50°

#### 11. m∠KMT = ?
→ Same as above → 50°

✔ Answer: 50°

#### 12. m∠NMT = ?
→ Inscribed angle at M intercepting arc NT
→ m∠NMT = ½ × 60° = 30°

✔ Answer: 30°

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Summary of Answers (Assumed Based on Standard Problems):

| Problem | Answer |
|--------|--------|
| 1 | Chords: AB, BC, AC (example) |
| 2 | Central angles: ∠AOB, ∠BOC, ∠AOC |
| 3 | Inscribed angles: ∠ABC, ∠BAC, ∠ACB |
| 4 | m∠COB = arc CB; m∠CAB = ½ arc CB |
| 5 | m∠EBD = ½ arc ED; m∠EFD = ½ arc ED |
| 6 | m∠CED = ½ arc CD |
| 7 | m∠POQ = arc PQ; m∠PQO = (180° − arc PQ)/2 |
| 8 | m∠CED = ½ arc CD; m∠CEA = ½ arc CA |
| 9 | m∠POQ = arc PQ; m∠PQO = (180° − arc PQ)/2 |
| 10 | m∠M = 50° (example) |
| 11 | m∠KMT = 50° |
| 12 | m∠NMT = 30° |

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To Get Exact Answers:

Please provide the diagram details or describe:

- The positions of points A, B, C, etc.
- Any given arc or angle measures
- Whether any lines are diameters, tangents, or chords
- Specific arc measures shown

With that, I can give precise answers.

But based on typical textbook problems, the above reasoning applies.

Let me know if you can describe the diagram!