$$
\begin{array}{l}
\text { 2. In } \triangle A B C \text {, } \angle C-45^{\circ} \text {, and } \\
\frac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}=\sqrt{3} \text {. }
\end{array}
$$

Then the degree measure of the largest interior angle is
$2.75^{\circ}$.
Let's assume $\angle A$ is the largest angle.
Since $\angle C=45^{\circ}$, we have $\angle A+\angle B=135^{\circ}$.

Thus,
$$
\begin{array}{l} 
\frac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}=\sqrt{3} \\
\Leftrightarrow(\sin A-\sqrt{3} \cos A)+(\sin B-\sqrt{3} \cos B) \\
\quad(\sin C-\sqrt{3} \cos C)=0 \\
\Leftrightarrow \sin \left(A-60^{\circ}\right)+\sin \left(B-60^{\circ}\right)=\sin 15^{\circ} \\
\Leftrightarrow 2 \sin \left(\frac{A-B}{2}-60^{\circ}\right) \cos \cdot \frac{A-B}{2}=\sin 15^{\circ} \\
\Leftrightarrow \cos \frac{A-B}{2}=\frac{\sin 15^{\circ}}{2 \sin \frac{15}{2}}=\cos \frac{15^{\circ}}{2} . \\
\therefore \angle A-\angle B=15^{\circ} .
\end{array}
$$

Since $\angle A+\angle B=135^{\circ}$, we have $\angle A=75^{\circ}$.