$7 \cdot 60$ In $\triangle A B C$, it is known that $\angle A=45^{\circ}, \angle B=60^{\circ}, B C=1$, then the area of $\triangle A B C$ is
(A) $\frac{1+\sqrt{3}}{8}$.
(B) $\frac{2+\sqrt{3}}{8}$.
(C) $\frac{1+\sqrt{3}}{4}$.
(D) $\frac{3+\sqrt{3}}{8}$.
(E) $\frac{\sqrt{6}+\sqrt{3}}{8}$.
(China Partial Provinces and Cities Junior High School Mathematics Correspondence Competition, 1986)
[Solution] Draw $B D \perp A C$, with $D$ as the foot of the perpendicular. In $\triangle A B C$, by the Law of Sines, we have
$$
A C=\frac{\sqrt{3}}{2} \cdot \frac{2}{\sqrt{2}}=\frac{\sqrt{6}}{2} \text {. }
$$

In the right triangle $\triangle B D C$,
$$
B D=\sin 75^{\circ}=\cos 15^{\circ}=\frac{\sqrt{2+\sqrt{3}}}{2} .
$$
$$
\begin{aligned}
\therefore \quad S_{\triangle A B C} & =\frac{1}{2} \cdot \frac{\sqrt{6}}{2} \cdot \frac{\sqrt{2+\sqrt{3}}}{2} \\
& =\frac{\sqrt{12+6 \sqrt{3}}}{8} \\
& =\frac{\sqrt{3(4+2 \sqrt{3})}}{8} \\
& =\frac{\sqrt{3(\sqrt{3}+1)^{2}}}{8}=\frac{\sqrt{3}(\sqrt{3}+1)}{8} \\
& =\frac{3+\sqrt{3}}{8} .
\end{aligned}
$$

Therefore, the correct choice is $(D)$.