如图已知角aob=a,在射线oa、ob上分别取点oa1=ob1,连接a1b1,在b1a1/b1b上分别取点a2b2,使b1b2=b1a2,连接a2b2.按此规律下去,记角a2b1b2=o1,角a3b2b3=o2,.,角An+1BnBn+1=On,则O1=,On=

Question

如图已知角aob=a,在射线oa、ob上分别取点oa1=ob1,连接a1b1,在b1a1/b1b上分别取点a2b2,使b1b2=b1a2,连接a2b2.按此规律下去,记角a2b1b2=o1,角a3b2b3=o2,.,角An+1BnBn+1=On,则O1= ,On=

Answer 1

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