19.47 A regular $n$-pointed star, it is a simple closed regular polygon with all sides of length $2 n$ (the figure below shows a part of it). Angles $A_{1}, A_{2}, \cdots, A_{n}$ are equal, and $B_{1}, B_{2}, \cdots, B_{n}$ are equal. If
$$
\angle A_{1}=\angle B_{1}-10^{\circ} \text {, }
$$

then $n$ equals
(A) 12.
(B) 18.
(C) 24.
(D) 36.
(E) 60.
(43rd American High School Mathematics Examination, 1992)
[Solution] Connect $A_{1} A_{2}, B_{n} B_{1}$, and $B_{1} B_{2}$.
Since the sum of the interior angles of the convex $n$-sided polygon $B_{1} B_{2} \cdots B_{n}$ is $(n-2) 180^{\circ}$, then
$$
\angle B_{n} B_{1} B_{2}=\frac{(n-2) 180^{\circ}}{n} \text {. }
$$

Let $\angle A_{1} B_{n} B_{1}=\alpha$,
we have $\alpha+\angle B_{1}+\alpha+\frac{(n-2) 180^{\circ}}{n}=360^{\circ}$,
which means $\alpha+\left(\angle A_{1}+10^{\circ}\right)+\alpha+\frac{(n-2) 180^{\circ}}{n}=360^{\circ}$,
and since $\alpha+\alpha+\angle A_{1}=180^{\circ}$, then
$$
\frac{(n-2) 180^{\circ}}{n}=360^{\circ}-190^{\circ}=170^{\circ} \text {, }
$$

which implies $(n-2) \cdot 180^{\circ}=n \cdot 170^{\circ}$,
or $18 n-36=17 n$, yielding $n=36$.
Therefore, the answer is $(D)$.