10. In the complex plane, take any three different roots of the equation $z^{100}-1=0$ as vertices to form a triangle. Then the number of different acute triangles is $\qquad$.

Translating the text into English while preserving the original formatting and line breaks, the result is as follows:
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10. In the complex plane, take any three different roots of the equation z^100 - 1 = 0 as vertices to form a triangle. Then the number of different acute triangles is      .
```
10. 39200.

It is known that the roots of $z^{100}-1=0$ lie on the unit circle, and the arc length between any two roots is $\frac{2 \pi}{100}$, which means the unit circle is evenly divided into 100 small arcs.

As shown in Figure 5, select any point $A$ as a vertex of the triangle, with 100 possible choices. Select vertices $B$ and $C$ in a clockwise direction. Let the arcs $\overparen{A B}$, $\overparen{C B}$, and $\overparen{A C}$ have $x$, $y$, and $z$ segments of small arcs, respectively.
The equivalent condition for $\triangle A B C$ to be an acute triangle is
$$
\left\{\begin{array}{l}
x + y + z = 100, \\
1 \leqslant x, y, z \leqslant 49
\end{array} \Rightarrow \left\{\begin{array}{l}
x+y+z=97, \\
0 \leqslant x, y, z \leqslant 48 .
\end{array}\right.\right.
$$

Next, we calculate the number of integer solutions to the system of equations (1).
$$
\begin{array}{l}
\text { Let } P_{1}=\{x \mid x+y+z=97, x \geqslant 49\}, \\
P_{2}=\{y \mid x+y+z=97, y \geqslant 49\}, \\
P_{3}=\{z \mid x+y+z=97, z \geqslant 49\}, \\
S=\{(x, y, z) \mid x+y+z=97, x, y, z \geqslant 0\}, \\
\left|\bar{P}_{1} \cap \bar{P}_{2} \cap \bar{P}_{3}\right|=|S|-\left|P_{1} \cup P_{2} \cup P_{3}\right| \\
=\mathrm{C}_{99}^{2}-\left(\left|P_{1}\right|+\left|P_{2}\right|+\left|P_{3}\right|-\right. \\
\left.\quad \sum_{i<j}\left|P_{i} \cap P_{j}\right|+\left|P_{1} \cap P_{2} \cap P_{3}\right|\right) \\
=\mathrm{C}_{99}^{2}-3 \mathrm{C}_{50}^{2}=1176 .
\end{array}
$$

Since it is counted three times, the number of acute triangles is
$$
\frac{100 \times 1176}{3}=39200 .
$$