Geometry Unit 10 - Circle Arcs Central 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 problem. However, based on the text you've provided, it appears to be a worksheet titled "Practice – Circle, Central, and Inscribed Angles", with several problems involving circles, central angles, inscribed angles, and arc measures.

Let me go through each problem based on the text you’ve shared and solve them using standard circle geometry rules:

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🔷 Key Rules Used:

1. Central Angle Theorem: The measure of a central angle is equal to the measure of its intercepted arc.
2. Inscribed Angle Theorem: The measure of an inscribed angle is half the measure of its intercepted arc.
3. Arc Addition Postulate: The total measure of a circle is 360°.
4. Angle in a Semicircle: An angle inscribed in a semicircle is a right angle (90°).
5. Tangent-Radius Perpendicularity: A tangent to a circle is perpendicular to the radius at the point of contact.

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Now let's solve each problem step-by-step:

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1. Find $ m\angle DAP $ if $ \angle DAP = 72^\circ $

This seems like a typo or misstatement — it says “find $ m\angle DAP $” but then gives $ \angle DAP = 72^\circ $.
Possibly this is just asking to identify or confirm that the angle is 72°.

✔ Answer: $ m\angle DAP = 72^\circ $

> Note: If there’s a diagram showing this as a central or inscribed angle, more context would help, but based on what's written, we assume it's given.

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2. Find $ m\angle ABC $ if $ m\angle ABC = 35^\circ $

Again, this appears to be giving the value. Possibly the question is to justify or use it in a larger problem.

But likely, it's a typo or formatting issue — perhaps it should say:
“Find $ m\angle ABC $ if arc AC = 70°” (since inscribed angle is half the arc).

But since it says $ m\angle ABC = 35^\circ $, and asks to find it...

✔ Answer: $ m\angle ABC = 35^\circ $

> This might be a placeholder — unless more context is available, we take it as given.

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3. Find $ m\angle BAC $ if $ m\angle BAC = 34^\circ $

Same pattern — possibly a typo. But again, it says "find" and gives the value.

Wait — maybe the actual problem is missing? Let's suppose the intended meaning is:

> "Given arc BC = 68°, find $ m\angle BAC $" — because inscribed angle intercepting arc BC would be half of that.

So if arc BC = 68°, then:
$$
m\angle BAC = \frac{1}{2} \times 68^\circ = 34^\circ
$$

✔ So, likely: $ m\angle BAC = 34^\circ $

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4. Find $ m\angle T $

From the description, this looks like a tangent and radius forming a right angle.

If $ \angle T $ is formed by a tangent and radius, then:
$$
m\angle T = 90^\circ
$$

✔ Answer: $ m\angle T = 90^\circ $

> Because tangent is perpendicular to radius at point of contact.

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5. $ TP $ is a diameter. Find $ m\angle T $

Assuming $ TP $ is a diameter, and $ \angle T $ is an inscribed angle on the semicircle.

But without more points, we need to infer.

Suppose triangle $ TQP $ has $ TP $ as diameter and $ Q $ on the circle → then $ \angle TQP = 90^\circ $ (Thales’ theorem).

But it asks for $ m\angle T $. That depends on where $ \angle T $ is.

Alternatively, if $ \angle T $ is at the center, and $ TP $ is diameter, then it may be a straight line — so angle could be 180°?

But likely, it's asking for an inscribed angle subtended by the diameter.

Wait — if $ \angle T $ is at the circumference and subtends the diameter, then it’s 90°.

But again, labeling matters.

But if $ TP $ is a diameter and $ \angle T $ is formed at point $ T $, between two chords, it could vary.

Without diagram, best assumption: if $ \angle T $ is an inscribed angle over the semicircle, then:

✔ Answer: $ m\angle T = 90^\circ $

(Using Thales' Theorem)

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6. Find $ m\angle G $

Looking at typical setups: if $ G $ is an inscribed angle intercepting arc $ EF $, and arc $ EF = x $, then $ m\angle G = \frac{1}{2}x $.

But no values are given. Wait — perhaps from previous info?

Maybe the diagram shows arcs labeled.

But based on common problems, if arc $ ED = 100^\circ $, and $ \angle G $ intercepts it, then:

$ m\angle G = \frac{1}{2} \times 100^\circ = 50^\circ $

But since nothing is specified, let’s suppose a common case.

Alternatively, if $ \angle G $ is a central angle, then it equals its arc.

But again, no data.

Wait — perhaps the original image had labels.

Let’s move to next ones.

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7. If $ m\angle ABC = 104^\circ $, find $ m\angle AOC $

Here, $ \angle ABC $ is an inscribed angle, and $ \angle AOC $ is a central angle, both intercepting the same arc $ AC $.

By the inscribed angle theorem:
$$
m\angle ABC = \frac{1}{2} \times m\angle AOC
$$
$$
104^\circ = \frac{1}{2} \times m\angle AOC
$$
$$
m\angle AOC = 2 \times 104^\circ = 208^\circ
$$

✔ Answer: $ m\angle AOC = 208^\circ $

> Note: This implies arc $ AC = 208^\circ $, which is possible (major arc).

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8. If $ m\angle ABC = 85^\circ $, find $ m\angle ADC $

Both $ \angle ABC $ and $ \angle ADC $ intercept the same arc $ AC $, assuming points $ A, B, C, D $ lie on the circle.

Then, angles subtended by the same arc are equal (if on the same side), or supplementary (if on opposite sides).

But typically, if both are inscribed angles intercepting arc $ AC $, they should be equal.

So:
$$
m\angle ADC = m\angle ABC = 85^\circ
$$

But wait — sometimes one is on the major arc, one on minor.

If $ \angle ABC $ and $ \angle ADC $ are on opposite sides of chord $ AC $, then they are supplementary.

But unless stated, assume same arc → equal.

But let’s check: if $ \angle ABC = 85^\circ $, and $ \angle ADC $ is on the other side, then:

Sum of opposite angles in cyclic quadrilateral = 180°.

So if $ ABCD $ is a cyclic quadrilateral, then:
$$
\angle ABC + \angle ADC = 180^\circ
$$
$$
85^\circ + m\angle ADC = 180^\circ
\Rightarrow m\angle ADC = 95^\circ
$$

✔ Answer: $ m\angle ADC = 95^\circ $

> Assuming $ ABCD $ is a cyclic quadrilateral, which is typical in such problems.

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## ✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $ m\angle DAP = 72^\circ $ |
| 2 | $ m\angle ABC = 35^\circ $ |
| 3 | $ m\angle BAC = 34^\circ $ |
| 4 | $ m\angle T = 90^\circ $ (tangent-radius) |
| 5 | $ m\angle T = 90^\circ $ (Thales' Theorem) |
| 6 | Insufficient info — likely $ m\angle G = \frac{1}{2} \text{arc} $ — need diagram |
| 7 | $ m\angle AOC = 208^\circ $ |
| 8 | $ m\angle ADC = 95^\circ $ |

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📌 Important Notes:

- Problems 1–3 seem to have answers given in the prompt — possibly typos or misformatting.
- Problem 6 lacks sufficient data without the diagram.
- Problems 7 and 8 are well-defined and solvable with standard theorems.

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If you can describe the diagrams or upload the image again (or paste the text clearly), I can give exact solutions with proper reasoning and diagrams.

Would you like me to create a visual explanation or walk through one of these problems with a sketch description?