$4 \cdot 223$ The sides of a triangle form an arithmetic sequence with a common difference of $d$, and the area of the triangle is equal to $S$. Find the lengths of the sides and the angles of the triangle. For the special case where $d=1, S=6$, solve the problem.
[Solution] Represent the sides of the triangle as $a=b-d, b, c=b+d(0<d<b)$, and substitute the expressions
$$
\begin{array}{l}
p=\frac{a+b+c}{2}=\frac{3 b}{2}, p-a=\frac{b}{2}+d, \\
p-b=\frac{b}{2}, p-c=\frac{b}{2}-d,
\end{array}
$$

into Heron's formula
$$
S^{2}=p(p-a)(p-b)(p-c) .
$$

This yields $S^{2}=\frac{3 b^{2}}{4}\left(\frac{b^{2}}{4}-d^{2}\right)$. Treating this as a quadratic equation in $b^{2}$, we solve to get
$$
b^{2}=2\left(d^{2}+\sqrt{d^{4}+\frac{4 s^{2}}{3}}\right) \text {, }
$$

so $b=\sqrt{2\left(d^{2}+\sqrt{d^{4}+\frac{4 s^{2}}{3}}\right)}$,
$$
\begin{array}{l}
a=b-d, \\
c=b+d .
\end{array}
$$
The angles $\alpha$ and $\beta$ opposite to $a$ and $b$ must be acute. Using
$$
S=\frac{b c}{2} \sin \alpha, S=\frac{a c}{2} \sin \beta, \gamma=180^{\circ}-(\alpha+\beta) \text {, we can solve for } \alpha, \beta \text {, and }
$$
$\gamma$.
If $d=1, s=6$, then $b=4, a=3, c=5$, and $\sin \alpha=\frac{3}{5}$, $\sin \beta=\frac{4}{5}$, yielding $\alpha=36^{\circ} 52^{\prime}, \beta=90^{\circ}-\alpha=53^{\circ} 8^{\prime}, \gamma=90^{\circ}$.