7. Let $\triangle A B C$ have three interior angles $\angle A, \angle B, \angle C$ with corresponding side lengths $a, b, c$. If $a<b<c$, and 
\[
\left\{\begin{array}{l}
\frac{b}{a}=\frac{\left|b^{2}+c^{2}-a^{2}\right|}{b c} \\
\frac{c}{b}=\frac{\left|c^{2}+a^{2}-b^{2}\right|}{c a} \\
\frac{a}{c}=\frac{\left|a^{2}+b^{2}-c^{2}\right|}{a b}
\end{array}\right.,
\]
then the ratio of the radian measures of $\angle A, \angle B, \angle C$ is $\qquad$
7. 1: 2: 4 Explanation: From $\frac{b}{a}=\frac{\left|b^{2}+c^{2}-a^{2}\right|}{b c}$, we get $\frac{\sin B}{\sin A}=2|\cos A| \Rightarrow \sin B=|\sin 2 A|$. Since $a<b<c$, we have $0<A<B<\frac{\pi}{2}$, so $B=2 A$. Similarly, from $\frac{c}{b}=\frac{\left|c^{2}+a^{2}-b^{2}\right|}{c a}$, we get $C=2 B$. Combining $A+B+C=\pi$, we have $A=\frac{\pi}{7}$, $B=\frac{2 \pi}{7}$, $C=\frac{4 \pi}{7}$. Therefore, the ratio of the radian measures of $\angle A, \angle B, \angle C$ is $1: 2: 4$.