2. Given in $\triangle A B C$, $\angle A: \angle B: \angle C=1$ : $2: 5$. Then $\frac{A B}{B C}=$ $\qquad$ .
$2.1+\sqrt{2}$.
Since $\angle A: \angle B: \angle C=1: 2: 5$, and $\angle A+\angle B+\angle C=180^{\circ}$, therefore, $\angle A=22.5^{\circ}, \angle B=45^{\circ}$.

As shown in Figure 8, draw $C D \perp A B$ at point $D$, and intercept $D E=B D$ on $A D$, connect $C E$.
We have $B D=C D$, then
$$
B C=\sqrt{2} B D \text {. }
$$

By the theorem of the three lines of an isosceles triangle coinciding, we know
$$
\begin{array}{l}
\angle C E D=\angle B=45^{\circ}, C E=B C=\sqrt{2} B D \\
\angle A C E=\angle A, A E=C E=\sqrt{2} B D
\end{array}
$$

Thus $A B=A E+D E+B D=(\sqrt{2}+2) B D$.
$$
\text { Hence } \frac{A B}{B C}=\frac{(\sqrt{2}+2) B D}{\sqrt{2} B D}=1+\sqrt{2} \text {. }
$$