$1992$ Ahsme Problem $20$: Part of an "n-pointed regular star" is shown. It is a simple closed polygon in which all $2n$ edges are congruent, angles $A_1,A_2,\cdots,A_n$ are congruent, and angles $B_1,B_2,\cdots,B_n$ are congruent. If the acute angle at $A_1$ is $10^\circ$ less than the acute angle at $B_1$, then $n=$
$\text{(A) } 12\quad \text{(B) } 18\quad \text{(C) } 24\quad \text{(D) } 36\quad \text{(E) } 60$
I am not sure how to solve it, and the solution meantions an opposing angle theorem (i am not sure what that is).
We have $B-A=10$. Let $B'=360-B$ (the internal angle at the $B$ points), the shape is $2n-$gon and its angles will add up to $180(2n-2)$. So we have \begin{eqnarray*} B-A&=&10 \\ n(A+B')&=& 180(2n-2). \end{eqnarray*} So ...