3. Given in $\triangle A B C$, $\angle A C B=90^{\circ}$, $\angle A B C=15^{\circ}, B C=1$. Then the length of $A C$ is ( ).
(A) $2+\sqrt{3}$
(B) $2-\sqrt{3}$
(C) 0.3
(D) $\sqrt{3}-\sqrt{2}$
3. (B).

As shown in Figure 4, construct $\angle E A \dot{D}=15^{\circ}$ intersecting $12 \mathrm{C}$ at $D$. Then $A D=$ $B D, \angle A D C=30^{\circ}$.
Let $A C=x$, then
$$
\begin{array}{l}
C D=\sqrt{3} x, \\
A D=2 x .
\end{array}
$$

Thus, $(2+\sqrt{3}) x=1$. Solving for $x$ yields $x=2-\sqrt{3}$.