10. Let the three equations concerning $x$ be
$$
\begin{array}{l}
x^{2}+2 x \sin A_{1}+\sin A_{2}=0, \\
x^{2}+2 x \sin A_{2}+\sin A_{3}=0, \\
x^{2}+2 x \sin A_{3}+\sin A_{1}=0
\end{array}
$$

all have real roots, $\angle A_{1}, \angle A_{2}, \angle A_{3}$ are three interior angles of a convex $(4 n+2)$-sided polygon $A_{1} A_{2} \cdots A_{4 n+2}$, and all interior angles are multiples of $30^{\circ}$. Then the sum of the interior angles of this convex $(4 n+2)$-sided polygon is $\qquad$.
$10.4 \pi$.
Since the interior angles of a convex $4 n+2$-sided polygon are all multiples of $30^{\circ}$, its interior angles can only be $30^{\circ}, 60^{\circ}, 90^{\circ}, 120^{\circ}, 150^{\circ}$, and the values of their sines can only be $\frac{1}{2}, \frac{\sqrt{3}}{2}, 1$. Also, because the three equations all have real roots, we have
$$
\begin{array}{l}
4 \sin ^{2} A_{1}-4 \sin A_{2} \geqslant 0,4 \sin ^{2} A_{2}-4 \sin A_{3} \geqslant 0, \\
4 \sin ^{2} A_{3}-4 \sin A_{1} \geqslant 0 .
\end{array}
$$

Therefore, $\sin A_{1}=\sin A_{2}=\sin A_{3}=1$.
Thus, $\angle A_{1}=\angle A_{2}=\angle A_{3}=90^{\circ}$.
According to the exterior angle sum theorem, the sum of the remaining $4 n-1$ exterior angles of this convex $4 n+2$-sided polygon is $90^{\circ}$. Therefore, each of these exterior angles can only be $30^{\circ}$. Hence, this polygon is a hexagon, and its interior angle sum is $4 \pi$.