6. Given that $O$ is a point inside the equilateral $\triangle A B C$, the ratio of the angles $\angle A O B$, $\angle B O C$, and $\angle A O C$ is 6:5:4. Then, in the triangle formed by $O A$, $O B$, and $O C$, the ratio of the angles opposite these sides is
$6.5: 3: 7$.
As shown in Figure 7, from $\angle A O B$:
$\angle B O C: \angle A O C=6: 5: 4$.
We know
$$
\begin{array}{l}
\angle A O B=144^{\circ}, \\
\angle B O C=120^{\circ}, \\
\angle A O C=96^{\circ} .
\end{array}
$$

With point $A$ as the center, rotate $\triangle A O B$ counterclockwise by $60^{\circ}$ to get $\triangle A O^{\prime} C$. Then,
$\triangle A O^{\prime} C \cong \triangle A O B$.
Therefore, $O^{\prime} C=O B$.
Connect $O O^{\prime}$, and we know $\triangle A O O^{\prime}$ is an equilateral triangle, so $O O^{\prime}=O A$. Thus, $\triangle O O^{\prime} C$ is the triangle formed by the sides $O A$, $O B$, and $O C$.

Since $\angle A O^{\prime} C=\angle A O B=144^{\circ}$ and $\angle A O^{\prime} O=60^{\circ}$, we have $\angle O O^{\prime} C=84^{\circ}$.
Also, $\angle A O C=96^{\circ}$, so $\angle O^{\prime} O C=36^{\circ}$. Therefore,
$\angle O C O^{\prime}=60^{\circ}$.
Hence, the ratio of the corresponding angles of the triangle formed by the sides $O A$, $O B$, and $O C$ is $60^{\circ}: 36^{\circ}: 84^{\circ}=5: 3: 7$.