初中数学《锐 角的三角函数值》教案
[07-25 15:10:09] 来源:http://www.89xue.com 数学知识大全 阅读:9531次
摘要:∴BB'+ DD'= 2OO'∴CC'- AA'= BB'+ DD'(2)如图,仿(1)证法可得CC'- AA'= 2OESinαDD'-BB = 2OFSinβ∵OESinα= OFSinβ,∴CC'- AA'= DD'- BB'证法二:(1)延长CB交MN于E,设AD与MN交于F, 又设∠AFA'= α,则∠BEB'= α,在Rt△EBB'中,∵BE= CE- CB∴BB'= BESinα- CBSinα 在R t△ECC'中,Sin&alph。
初中数学《锐 角的三角函数值》教案,标签:数学,http://www.89xue.com
∴BB'+ DD'= 2OO'
∴CC'- AA'= BB'+ DD'
(2)如图,仿(1)证法可得
CC'- AA'= 2OESinα
DD'-BB = 2OFSinβ
∵OESinα= OFSinβ,
∴CC'- AA'= DD'- BB'
证法二:(1)延长CB交MN于E,设AD与MN交于F, 又设∠AFA'= α,则∠BEB'= α,在Rt△EBB'中,
∵BE= CE- CB
∴BB'= BESinα- CBSinα
在R t△ECC'中,Sinα= ,
∴CC’= CESinα
∵CC'- BB'= BCSinα
在Rt△AA'F与Rt△FDD'中.
AA'= AFSinα, DD'= DFSinα
∵DF= AD - AF
∴DD'= ADSinα- AFSinA'
∴DD'= ADSinα- AA'
∴DD'+ AA'= ADSinα
∵AD= BC, ∴CC'- BB'= DD'+ AA'
∴CC'- AA'= BB'+ DD'
(2)仿证法(1)同样可证得
CC'+ BB'= BCSinα
AA'+ DD'= ADSinα
∴CC'+ BB'= AA'+DD',
∴CC'- AA'= DD'- BB'
证法三:(1)如图,作DE⊥CC', 则DD'C'E为矩形,∴CE= CC'- DD'
设∠AFA'= α, 则易知∠CDE= α 在Rt△CDE中,
∴CC'- DD'= CDSinα
在Rt△AFA'中, AA'= AFSinα
在Rt△FBB'中, BB'= BFSinα
∴BB'= (AB- AF)Sinα= ABSinα- AFSinα
∴AA'+ BB'= ABSinα
∵AB = CD, ∵AA'+ BB'= CC'- DD'
∴CC'- AA'= DD'+ BB'
(2)如图,仿(1)同法可证:
CC'- AA'= DD'-BB'
【创新园地】
已知△ABC中,∠BAC= 120°,∠ABC=15°,
∠A,∠B,∠C的对边分别为a, b ,c那么a:b:c = _________ (本结论中不含任何三角函数,但保留根号,请考虑多种解法).
解法一:过点B作BD⊥AC交CA的延长线于点D.
∴∠BAC=120°,
∠ABC= 15°, ∴∠ACB= ∠DBC=45°,∠ABD= 30°
在Rt△ABD中,Sin30°= ∴AD= c
Cos30°= , ∴BD =
∴b - BD - AD =
a =
∴ a:b:c =
=
解法二:如图,作AD⊥BC, 交BC于D,在AB上取AE = AC, 连CE, 作AF⊥CE,交CE于F,则∠ACE = ∠AEC= ,(此括号内不是文章内容,来自学习方法网,阅读请跳过),∠BCE= ∠ACB- 30°= 45°- 30° = 15°
∴ △BEC为等腰三角形,∴BE= CE
设AD = CD = 1, 则AC = , 即b =
∴CE = 2 AC Cos30°=
∴AB= AE + EB = + , 即c = +
∴BD =
∴BC = BD + DC = 3 + ,即a = 3 +
∴ a:b:c = (3+ ): :( + )
=
解法三:如图,作AD⊥BC, 交BC于D, 在BC上取点E,使∠BAE = ∠B = 15°,那么,连接AE, 得:∠AEC = 30°,AE = BE. 设AD = DC = 1, 则AC = ,即b = ,AE= BE = 2AD = 2,DE = AE•Cos30° =
∴
即c = +
∴ a:b:c = (3+ ) : :( + )
=
解法四:如图,BD = x, 则2x2 = a2,
∴x =
∴BB'+ DD'= 2OO'
∴CC'- AA'= BB'+ DD'
(2)如图,仿(1)证法可得
CC'- AA'= 2OESinα
DD'-BB = 2OFSinβ
∵OESinα= OFSinβ,
∴CC'- AA'= DD'- BB'
证法二:(1)延长CB交MN于E,设AD与MN交于F, 又设∠AFA'= α,则∠BEB'= α,在Rt△EBB'中,
∵BE= CE- CB
∴BB'= BESinα- CBSinα
在R t△ECC'中,Sinα= ,
∴CC’= CESinα
∵CC'- BB'= BCSinα
在Rt△AA'F与Rt△FDD'中.
AA'= AFSinα, DD'= DFSinα
∵DF= AD - AF
∴DD'= ADSinα- AFSinA'
∴DD'= ADSinα- AA'
∴DD'+ AA'= ADSinα
∵AD= BC, ∴CC'- BB'= DD'+ AA'
∴CC'- AA'= BB'+ DD'
(2)仿证法(1)同样可证得
CC'+ BB'= BCSinα
AA'+ DD'= ADSinα
∴CC'+ BB'= AA'+DD',
∴CC'- AA'= DD'- BB'
证法三:(1)如图,作DE⊥CC', 则DD'C'E为矩形,∴CE= CC'- DD'
设∠AFA'= α, 则易知∠CDE= α 在Rt△CDE中,
∴CC'- DD'= CDSinα
在Rt△AFA'中, AA'= AFSinα
在Rt△FBB'中, BB'= BFSinα
∴BB'= (AB- AF)Sinα= ABSinα- AFSinα
∴AA'+ BB'= ABSinα
∵AB = CD, ∵AA'+ BB'= CC'- DD'
∴CC'- AA'= DD'+ BB'
(2)如图,仿(1)同法可证:
CC'- AA'= DD'-BB'
【创新园地】
已知△ABC中,∠BAC= 120°,∠ABC=15°,
∠A,∠B,∠C的对边分别为a, b ,c那么a:b:c = _________ (本结论中不含任何三角函数,但保留根号,请考虑多种解法).
解法一:过点B作BD⊥AC交CA的延长线于点D.
∴∠BAC=120°,
∠ABC= 15°, ∴∠ACB= ∠DBC=45°,∠ABD= 30°
在Rt△ABD中,Sin30°= ∴AD= c
Cos30°= , ∴BD =
∴b - BD - AD =
a =
∴ a:b:c =
=
解法二:如图,作AD⊥BC, 交BC于D,在AB上取AE = AC, 连CE, 作AF⊥CE,交CE于F,则∠ACE = ∠AEC= ,(此括号内不是文章内容,来自学习方法网,阅读请跳过),∠BCE= ∠ACB- 30°= 45°- 30° = 15°
∴ △BEC为等腰三角形,∴BE= CE
设AD = CD = 1, 则AC = , 即b =
∴CE = 2 AC Cos30°=
∴AB= AE + EB = + , 即c = +
∴BD =
∴BC = BD + DC = 3 + ,即a = 3 +
∴ a:b:c = (3+ ): :( + )
=
解法三:如图,作AD⊥BC, 交BC于D, 在BC上取点E,使∠BAE = ∠B = 15°,那么,连接AE, 得:∠AEC = 30°,AE = BE. 设AD = DC = 1, 则AC = ,即b = ,AE= BE = 2AD = 2,DE = AE•Cos30° =
∴
即c = +
∴ a:b:c = (3+ ) : :( + )
=
解法四:如图,BD = x, 则2x2 = a2,
∴x =
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