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To express the point \((4, 4\sqrt{3}, 0)\) in polar coordinates, let's go through the following steps.

### Step 1: Identify the coordinates
The point is given in Cartesian coordinates \((x, y, z)\) as:
\[
x = 4, \quad y = 4\sqrt{3}, \quad z = 0
\]

### Step 2: Calculate \(r\)
In polar coordinates, \(r\) (the radial distance) is calculated as:
\[
r = \sqrt{x^2 + y^2 + z^2}
\]
Substitute \(x = 4\), \(y = 4\sqrt{3}\), and \(z = 0\):
\[
r = \sqrt{4^2 + (4\sqrt{3})^2 + 0^2}
= \sqrt{16 + 48}
= \sqrt{64}
= 8
\]

### Step 3: Calculate \(\theta\)
The angle \(\theta\) in polar coordinates is given by:
\[
\theta = \tan^{-1}\left(\frac{y}{x}\right)
\]
Substitute \(x = 4\) and \(y = 4\sqrt{3}\):
\[
\theta = \tan^{-1}\left(\frac{4\sqrt{3}}{4}\right)
= \tan^{-1}(\sqrt{3})
= \frac{\pi}{3} \text{ radians}
\]

### Step 4: Determine \(z\)-axis component (if cylindrical)
In this problem, since \(z = 0\), the point lies in the \(xy\)-plane. In cylindrical coordinates, the \(z\)-coordinate remains as it is. So \(z = 0\) remains the same in polar form.

### Final Answer
The polar coordinates (or cylindrical coordinates) for the point \((4, 4\sqrt{3}, 0)\) are:
\[
(r, \theta, z) = (8, \frac{\pi}{3}, 0)
\]

Would you like further details, or do you have any questions about this conversion?

### Related Questions
1. How would you convert a point with a non-zero \(z\)-coordinate to polar or cylindrical coordinates?
2. What are the steps to convert polar coordinates back to Cartesian coordinates?
3. How would the conversion change if the point were located in a different quadrant?
4. Can you express the same point in spherical coordinates?
5. How do you interpret negative radial distances in polar coordinates?

### Tip
Always double-check your angle calculation to ensure it's in the correct quadrant, as \(\tan^{-1}\) provides the principal value.

Express the point (4, 4√3, 0) in polar coordinates.

Learn how to convert the point (4, 4√3, 0) from Cartesian to polar coordinates. This detailed solution includes step-by-step calculations using trigonometry and the Pythagorean Theorem, making it suitable for students in Grades 11-12.

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