4. Let $I$ be the incenter of $\triangle A B C$, and
$$
3 \overrightarrow{I A}+4 \overrightarrow{I B}+5 \overrightarrow{I C}=0 \text {. }
$$

Then the size of $\angle C$ is $\qquad$
4. $\frac{\pi}{2}$.

Given that $I$ is the incenter of $\triangle ABC$, we have $a \overrightarrow{I A}+b \overrightarrow{I B}+c \overrightarrow{I C}=\mathbf{0}$. Also, $3 \overrightarrow{I A}+4 \overrightarrow{I B}+5 \overrightarrow{I C}=\mathbf{0}$, hence $a: b: c=3: 4: 5$.

Therefore, $\angle C=\frac{\pi}{2}$.