The angles $\alpha, \beta, \gamma$ of a triangle are in arithmetic progression. If $\sin 20\alpha$, $\sin 20\beta$, and $\sin 20\gamma$ are in arithmetic progression, how many different values can $\alpha$ take?

$ 
\textbf{(A)}\ 1
\qquad\textbf{(B)}\ 2
\qquad\textbf{(C)}\ 3
\qquad\textbf{(D)}\ 4
\qquad\textbf{(E)}\ \text{None of the above}
$
1. Given that the angles $\alpha, \beta, \gamma$ of a triangle are in arithmetic progression, we can write:
   \[
   \beta = \alpha + d \quad \text{and} \quad \gamma = \alpha + 2d
   \]
   Since the sum of the angles in a triangle is $\pi$, we have:
   \[
   \alpha + \beta + \gamma = \pi \implies \alpha + (\alpha + d) + (\alpha + 2d) = \pi \implies 3\alpha + 3d = \pi \implies \alpha + d = \frac{\pi}{3}
   \]
   Therefore, we can express $d$ as:
   \[
   d = \frac{\pi}{3} - \alpha
   \]

2. We are given that $\sin 20\alpha$, $\sin 20\beta$, and $\sin 20\gamma$ are in arithmetic progression. This implies:
   \[
   2\sin 20\beta = \sin 20\alpha + \sin 20\gamma
   \]
   Using the angle sum and difference identities, we can write:
   \[
   \sin 20\gamma = \sin 20(\alpha + 2d) = \sin (20\alpha + 40d)
   \]
   and
   \[
   \sin 20\beta = \sin 20(\alpha + d) = \sin (20\alpha + 20d)
   \]

3. Substituting these into the arithmetic progression condition, we get:
   \[
   2\sin (20\alpha + 20d) = \sin 20\alpha + \sin (20\alpha + 40d)
   \]
   Using the sum-to-product identities, we have:
   \[
   \sin 20\alpha + \sin (20\alpha + 40d) = 2 \sin \left( 20\alpha + 20d \right) \cos (20d)
   \]
   Therefore, the equation becomes:
   \[
   2\sin (20\alpha + 20d) = 2 \sin (20\alpha + 20d) \cos (20d)
   \]
   Dividing both sides by $2\sin (20\alpha + 20d)$ (assuming $\sin (20\alpha + 20d) \neq 0$), we get:
   \[
   1 = \cos (20d)
   \]
   This implies:
   \[
   \cos (20d) = 1 \implies 20d = 2k\pi \quad \text{for some integer } k
   \]
   Since $d = \frac{\pi}{3} - \alpha$, we have:
   \[
   20\left(\frac{\pi}{3} - \alpha\right) = 2k\pi \implies \frac{20\pi}{3} - 20\alpha = 2k\pi \implies 20\alpha = \frac{20\pi}{3} - 2k\pi \implies \alpha = \frac{\pi}{3} - \frac{k\pi}{10}
   \]

4. To ensure $\alpha$ is a valid angle in a triangle, it must satisfy $0 < \alpha < \frac{\pi}{3}$. Therefore:
   \[
   0 < \frac{\pi}{3} - \frac{k\pi}{10} < \frac{\pi}{3}
   \]
   Solving these inequalities:
   \[
   0 < \frac{\pi}{3} - \frac{k\pi}{10} \implies \frac{k\pi}{10} < \frac{\pi}{3} \implies k < \frac{10}{3} \implies k \leq 3
   \]
   and
   \[
   \frac{\pi}{3} - \frac{k\pi}{10} < \frac{\pi}{3} \implies -\frac{k\pi}{10} < 0 \implies k > 0
   \]
   Therefore, $k$ can be $1, 2, 3$.

5. Thus, the possible values of $\alpha$ are:
   \[
   \alpha = \frac{\pi}{3} - \frac{\pi}{10}, \quad \alpha = \frac{\pi}{3} - \frac{2\pi}{10}, \quad \alpha = \frac{\pi}{3} - \frac{3\pi}{10}
   \]

The final answer is $\boxed{3}$