4. In $\triangle A B C$, $\angle C=3 \angle A, a=27, c=48$. Then $b$ $=$ ـ $\qquad$
4. 35 .

As shown. Draw the trisectors of $\angle C$ as $C D$ and $C E$.
$$
\begin{array}{l}
\text { Let } C D=1, A N=x, \\
A E=E C=y .
\end{array}
$$

From $\triangle C D B \sim \triangle A C B$, we have
$$
\begin{array}{l}
\frac{a}{c}=\frac{t}{b}=\frac{c-x}{a}, \\
t=\frac{a b}{c}, x=\frac{c^{2}-a^{2}}{c} .
\end{array}
$$

Also, from $\triangle C E D \sim \triangle A C D$, we have
$$
\frac{t}{x}-\frac{y}{b}=\frac{x-y}{t} .
$$

Thus, $\frac{x}{b+t}=\frac{t}{x}, x^{2}=t(b+t)$.
(1) Substituting (1) into (2) and simplifying, we get
$$
b^{2}=\frac{1}{a}(c+a)(c-a)^{2} .
$$

Therefore, $b=35$.