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The problem provided involves finding the radius of a circle when a central angle of 12 degrees subtends an arc of length 22 cm. 

To solve this, we can use the formula for the length of an arc in a circle:

\[
s = r \theta
\]

Where:
- \( s \) is the arc length,
- \( r \) is the radius of the circle,
- \( \theta \) is the central angle in radians.

### Step 1: Convert the angle from degrees to radians.
We know that:
\[
\theta = \frac{\pi}{180} \times \text{angle in degrees}
\]
For 12 degrees:
\[
\theta = \frac{\pi}{180} \times 12 = \frac{\pi}{15} \, \text{radians}
\]

### Step 2: Substitute the values into the arc length formula.

The arc length is given as 22 cm, so:
\[
22 = r \times \frac{\pi}{15}
\]

### Step 3: Solve for the radius \( r \).

Rearranging the equation to solve for \( r \):
\[
r = \frac{22 \times 15}{\pi} = \frac{330}{\pi} \, \text{cm}
\]

Thus, the radius of the circle is \( \frac{330}{\pi} \) cm.

Would you like further details on any part of the solution?

Here are 5 questions related to this topic:
1. How do you convert degrees to radians?
2. What is the relationship between arc length, radius, and central angle in a circle?
3. How do you rearrange equations to solve for different variables?
4. What is the significance of using exact answers versus approximate values in geometry problems?
5. Can this formula be applied to find other properties of a circle (e.g., area or circumference)?

**Tip:** Always remember to convert angles to radians when dealing with formulas involving arc length or sectors of a circle.

A central angle of 12 degrees is subtended by an arc of length 22 cm. Find the radius of the circle. Write the exact answer but not the approximate value (For example, write 4/5 or 4/5 but not 0.8). Your answer should be written in terms of Pi or pi or π if there is a π in the answer.

Learn how to find the radius of a circle when a central angle of 12 degrees subtends an arc of length 22 cm. This problem involves converting degrees to radians and applying the arc length formula. Suitable for students in Grades 9-11.

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