设∠A 2 B 2 B 1 =y,
则θ 2 +y=180°①,θ 1 +2y=180°②,
①×2-②得:2θ2-θ 1 =180°,
∴θ 2 =
(180°+ θ 1)
2 ;
…
θ n =
(180°+ θ n-1)/2 .所以
θ n =
[( 2 n -1)180°+α]/
2 n
设∠A 2 B 2 B 1 =y,
则θ 2 +y=180°①,θ 1 +2y=180°②,
①×2-②得:2θ2-θ 1 =180°,
∴θ 2 =
(180°+ θ 1)
2 ;
…
θ n =
(180°+ θ n-1)/2 .所以
θ n =
[( 2 n -1)180°+α]/
2 n