如图,继续做角A1BC和角A1CD的平分线交于点A2
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/27 15:59:39
角A1=180-角A1BC-角A1CB=180-1/2角ABC-(角ACB+角ACA1)=180-1/2角ABC-角ACB-1/2角ACD=180-1/2角ABC-角ACB-1/2(180-角ACB)
∠A1=1/2∠A,∠An=(1/2)^n∠A
证:因三角形ABC绕点B顺时针旋转90度得到三角形A1BC1则:角CBC1=90度,又因角ACB=90度,AC=BC=10则:A1C1平行于CBBC=A1C1则:四边形CBA1C1是平行四边形
/>∵∠A+∠ABC+∠ACB=180∴∠ABC+∠ACB=180-∠A∵∠ACD=180-∠ACB,CA1平分∠ACD∴∠A1CD=∠ACD/2=(180-∠ACB)/2=90-∠ACB/2∵BA1
∠1不知道是什么,当是∠A=96好了.根据我画的图,∠A(i+1)CD=1/2∠AiCD=1/2(∠Ai+∠AiBC)∠A(i+1)=∠A(i+1)CD-∠A(i+1)BC而∠A(i+1)BC=1/2
∠A1=∠A1CD-∠A1BC/2=(∠A+∠ABC)/2-∠A1BC=∠A/2.同样可得∠A2=∠A/4;∠A3=∠A/8;∠A4=∠A/16;.∠An=∠A/2^n;再问:最后一个问题嘞?
∵∠ACD=∠A+∠ABC,CA1平分∠ACD∴∠DCA1=∠ACD/2=(∠A+∠ABC)/2=∠A/2+∠ABC/2∵BA1平分∠ABC∴∠DBA1=∠ABC/2∵∠DCA1=∠DBA1+∠A1∴
解题思路:先得出∠A1=1/2∠A,∠A2=1/4∠A,可得∠An=(1/2)^n∠A,再将∠A=32°,n=4,代入即可得∠A4的度数。解题过程:
△ABC中,∵∠A=∠ACD-∠ABC,A1是∠ABC角平分与∠ACD的平分线的交点,∠A=α,∴∠A1=∠A1CD-∠A1BC=1/2×(∠ACD-∠ABC)=1/2×∠A;同理可得,∠A2=1/2
角A2=角A的一半角A3=角A2的一半角A4=角A3的一半因此,角A4=角A的八分之一=a/8
∠AN=∠A/2=α/2^N,∠A=32,∠A4=32/2⁴=2
(1)∠A+1/2∠ABC=1/2∠ACD+∠Aι∠ACD=∠A+∠ABC∴∠Aι=1/2∠A(2)同理:∠An=(1/2)^n∠A(3)A4=(1/2)^4*∠A=64°/16=4°.
(1)∵A1B是∠ABC的平分线,A1C是∠ACD的平分线,∴∠A1BC=12∠ABC,∠A1CD=12∠ACD,又∵∠ACD=∠A+∠ABC,∠A1CD=∠A1BC+∠A1,∴12(∠A+∠ABC)
∵∠ACD=∠A+∠ABC,CA1平分∠ACD∴∠A1CD=∠ACD/2=(∠A+∠ABC)/2∵BA1平分∠ABC∴∠A1BC=∠ABC/2∴∠A1CD=∠A1+∠A1BC=∠A1+∠ABC/2∴∠
∵∠ACD=∠A+∠ABC,∠A1CD=∠A1+∠A1BC,∠ACD=2∠A1CD,∠ABC=2∠A1BC,∴2∠A1CD=∠A+2∠A1BC,即∠A1CD=12∠A+∠A1BC,∴∠A1=∠A2=α
OA=5,角AOB=30°,所以AB=5*tan30°,B点坐标为(5,5/√3)角A1OA=60°,OA1=OA=5,A1横坐标为5*cos60°=2.5,纵坐标为5*sin60°=2.5*√3,A
A列是不是最多是不是两位数如果是C1单元格公式可以写成=IF(LEN(A1)=1,SUBSTITUTE(B1,A1,),IF(LEN(A1)=2,SUBSTITUTE(SUBSTITUTE(B1,LE
以A和A1两个角为例,∠ACD=∠A+∠ABC,∠A1CD=1/2*∠ACD=1/2*∠A+1/2*∠ABC=1/2*∠A+∠A1BC,∠A1CD为外角=∠A1+∠A1BC所以,∠A1=1/2∠A,因
∵∠ACA1=∠A1CD=12∠ACD=12(∠A+∠ABC),又∵∠ABA1=∠A1BD=12∠ABD,∠A1CD=∠A1BD+∠A1,∴∠A1=12∠A=12α.同理∠A2=12∠A1,…即每次作