11. In a non-isosceles $\triangle ABC$, the sides opposite to angles $A, B, C$ are $a, b, c$ respectively, and it satisfies $(2c - b) \cos C = (2b - c) \cos B$.
(1) Find the size of angle $A$;
(2) If $a = 4$, find the range of the area of $\triangle ABC$.
$$
\begin{array}{l}
\text { (1) }(2 \sin C-\sin B) \cos C=(2 \sin B-\sin C) \cos B \\
\Rightarrow \sin 2 C-\sin 2 B=\sin (B-C) \Rightarrow 2 \cos (C+B) \sin (C-B)=\sin (B-C) \\
\Rightarrow 2 \cos (C+B)=-1 \Rightarrow \cos A=\frac{1}{2} \Rightarrow A=60^{\circ} \text {. }
\end{array}
$$
(2) From $a^{2}=b^{2}+c^{2}-b c>b c \Rightarrow 0<b c<16 \Rightarrow S_{\triangle A B C}=\frac{1}{2} b c \sin A<4 \sqrt{3}$, so the range of the area of $\triangle A B C$ is $(0,4 \sqrt{3})$.