30 Given that the three sides of $\triangle A B C$ are all integers, $\angle A=2 \angle B, \angle C>90^{\circ}$. Find the minimum perimeter of $\triangle A B C$.
30. Let the corresponding side lengths of $\triangle ABC$ be $a, b, c$. Draw the angle bisector of $\angle A$ intersecting $BC$ at point $D$, then
$$C D=\frac{a b}{b+c},$$

Using $\triangle A C D \backsim \triangle B C A$, we know
$$\begin{array}{c}
\frac{C D}{b}=\frac{b}{a} \\
a^{2}=b(b+c)
\end{array}$$

That is $\square$
And
Therefore
By
Let
$$\begin{array}{c}
\angle C>90^{\circ} \\
c^{2}>a^{2}+b^{2} \\
a^{2}=b(b+c) \\
(b, b+c)=d
\end{array}$$

Then $(b, c)=d$, and $d^{2} \mid a^{2}$, hence $d \mid a$.
To find the minimum value of $a+b+c$, we can set $d=1$, in which case both $b$ and $b+c$ are perfect squares. Let
$$b=m^{2}, b+c=n^{2}, m, n \in \mathbf{N}^{*}$$

Then $a=m n$. Using $a+b>c$ and $c^{2}>a^{2}+b^{2}$, we know

And $\square$
$$\begin{array}{c}
m n+m^{2}>n^{2}-m^{2} \\
\left(n^{2}-m^{2}\right)^{2}>(m n)^{2}+m^{4}
\end{array}$$

Thus
$$m>n-m$$

That is $\square$
$$n3 m^{2} n^{2},$$

That is $\square$
$$n^{2}>3 m^{2},$$

So $\square$
$$3 m^{2}<n^{2}<4 m^{2}$$

Thus, there is a perfect square between $3 m^{2}$ and $4 m^{2}$, which requires $m \geqslant 4$, at this time $n \geqslant 7$, hence
$$a+b+c \geqslant 4 \times 7+7^{2}=77$$

Clearly, $(a, b, c)=(28,16,33)$ satisfies the conditions, so the minimum perimeter of $\triangle A B C$ is 77.