1. Given that the three sides of a triangle are consecutive natural numbers. If the largest angle is twice the smallest angle, then the perimeter of the triangle is $\qquad$ .
-,1.15.
Assuming the three sides of a triangle are $n-1$, $n$, and $n+1$, with the largest angle being $2 \theta$ and the smallest angle being $\theta$.
Then, by the Law of Sines, we have
$$
\frac{n-1}{\sin \theta}=\frac{n+1}{\sin 2 \theta} \Rightarrow \cos \theta=\frac{n+1}{2(n-1)} \text {. }
$$

By the Law of Cosines, we get
$$
\begin{array}{l}
\cos \theta=\frac{(n+1)^{2}+n^{2}-(n-1)^{2}}{2 n(n+1)} \\
=\frac{n+4}{2(n+1)} .
\end{array}
$$

Thus, $\cos \theta=\frac{n+1}{2(n-1)}=\frac{n+4}{2(n+1)}$
$$
\Rightarrow n=5 \text {. }
$$

Therefore, the perimeter of the triangle is 15.