7. Given the equation of circle $\odot C$ is
$$
x^{2}+y^{2}-4 x-2 y+1=0 \text {, }
$$

the line $y=\left(\tan 10^{\circ}\right) x+\sqrt{2}$ intersects $\odot C$ at points
$A$ and $B$. The sum of the inclination angles of lines $A C$ and $B C$ is
$\qquad$
7. $200^{\circ}$.
The equation of $\odot C$ is $(x-2)^{2}+(y-1)^{2}=4$, the distance from the center $(2,1)$ to the line $y=\left(\tan 10^{\circ}\right) x+\sqrt{2}$ is
$$
d=\frac{\left|2 \tan 10^{\circ}-1+\sqrt{2}\right|}{\sqrt{1+\tan ^{2} 10^{\circ}}}<2,
$$

which means the given line intersects $\odot C$ at points $A$ and $B$ (as shown in Figure 2).
Let the inclination angle of line $B C$ be $\alpha$, and the inclination angle of line $A C$ be $\beta$. Then
$$
\begin{array}{l}
\alpha=10^{\circ}+\angle C B A, \\
\angle C A B=10^{\circ}+\left(180^{\circ}-\beta\right) .
\end{array}
$$

Since $\angle C B A=\angle C A B$, we have
$$
\begin{array}{l}
\alpha=10^{\circ}+10^{\circ}+\left(180^{\circ}-\beta\right) \\
\Rightarrow \alpha+\beta=200^{\circ} .
\end{array}
$$