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2021年正弦定理和余弦定理.pdf

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1、1 第三章三角函数、三角恒等变换及解三角形第7 课时正弦定理和余弦定理1.(必修 5P10习题 1.1 第 1(2)题改编)在ABC中，若A 60，B45，BC 32，则 AC _答案：23 解析：在 ABC中，ACsinBBCsinA，AC BC2 sinBsinA323223223.2.(必修 5P24复习题第1(2)题改编)在ABC中，a3，b 1，c2，则 A _答案：60解析：由余弦定理，得cosAb2c2a22bc1 432313212，0 A，A 60.3.(必修 5P17习题 1.2 第 6 题改编)在ABC中，a、b、c 分别为角A、B、C所对的边，若 a2bcosC，则此三

2、角形一定是_三角形答案：等腰解析：因为a2bcosC，所以由余弦定理得a2b2a2b2c22ab，整理得b2c2，故此三角形一定是等腰三角形4.(必修 5P17习题 6 改编)已知 ABC的三边长分别为a、b、c，且 a2b2c2ab，则C _答案：60解析：cosCa2b2 c22abab2ab12.0 C180，C 60.5.(必修 5P11习题 1.1 第 6(1)题改编)在ABC中，a32，b23，cosC13，则ABC的面积为 _答案：43 解析：cosC13，sinC 223，SABC12absinC 123 323 233223 43.1.正弦定理：asinAbsinBcsinC

3、2R(其中 R为ABC外接圆的半径)2.余弦定理a2 b2 c22bccosA，b2a2c22accosB；c2a2b22abcosC 或 cosAb2c2a22bc，精品w o r d 学习资料可编辑资料-精心整理-欢迎下载-第 1 页，共 8 页2 cosBa2c2b22ac，cosCa2b2c22ab.3.三角形中的常见结论(1)A BC.(2)在三角形中大边对大角，大角对大边：ABabsinAsinB.(3)任意两边之和大于第三边，任意两边之差小于第三边(4)ABC的面积公式 S12a2 h(h 表示 a 边上的高)；S12absinC 12acsinB 12bcsinA abc4R

4、；S12r(a bc)(r为内切圆半径)；SP（Pa）（Pb）（Pc），其中P12(a bc)备课札记题型 1 正弦定理解三角形例 1在ABC中，a3，b2，B45.求角 A、C和边 c.解：由正弦定理，得asinAbsinB，即3sinA2sin45，sinA32.ab，A 60或 A120.当 A60时，C180 45 60 75，cbsinCsinB622；当 A120时，C 180 45 120 15，cbsinCsinB622.变式训练在ABC中，(1)若 a4，B 30，C 105，则 b_(2)若 b3，c2，C45，则 a_(3)若 AB 3，BC 6，C30，则 A _精品

5、w o r d 学习资料可编辑资料-精心整理-欢迎下载-第 2 页，共 8 页文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4

6、I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4

7、D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6

8、B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M

9、5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY

10、7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10

11、R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A23 答案：(1)22(2)无解(3)45 或 135解析：(1)已知两角和一边只有一解，

12、由B30，C 105，得 A 45.由正弦定理，得basinBsinA4sin30 sin45 22.(2)由正弦定理得sinB bsinCC321，无解(3)由正弦定理BCsinAABsinC，得6sinA312，sinA 22.BCAB，AC，A45或 135.题型 2 余弦定理解三角形例 2在ABC中，a、b、c 分别是角A、B、C的对边，且cosBcosCb2ac.(1)求角 B的大小；(2)若 b13，ac4，求 ABC的面积解：(1)由余弦定理知：cosBa2c2b22ac，cosCa2b2c22ab.将上式代入cosBcosCb2ac，得a2 c2 b22ac22aba2b2c2

13、b2ac，整理得 a2c2b2ac.cosBa2c2b22acac2ac12.B 为三角形的内角，B 23.(2)将 b13，ac4，B 23代入b2a2c22accosB，得b2(a c)22ac2accosB，13 16 2ac 112，ac 3.SABC12acsinB 334.备选变式（教师专享）(20142南京期末)在ABC中，角 A、B、C所对的边分别是a、b、c，已知 c2，C3.(1)若ABC的面积等于3，求 a、b；(2)若 sinC sin(B A)2sin2A，求 ABC的面积解：(1)由余弦定理及已知条件，得a2b2ab4.因为 ABC的面积等于3，所以12absinC

14、 3，得 ab4.联立方程组a2b2ab4，ab4，解得 a2，b2.(2)由题意得sin(B A)sin(B A)4sinAcosA，所以 sinBcosA 2sinAcosA.精品w o r d 学习资料可编辑资料-精心整理-欢迎下载-第 3 页，共 8 页文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2

15、文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:C

16、K4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1

17、S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 H

18、T6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U

19、1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2

20、ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M

21、10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A24 当 cosA0 时，A2，所以 B6，所以 a433，b233.当 cosA0 时，得 sinB 2sinA，由正弦定理得b2a，联立方程组a2b2ab4，b2a，解得 a233，b433.所以 ABC的面积 S12absinC 233.题型 3 三角形形状的判定例 3在ABC中，a、b、c 分别表示三个内角 A、B、C 的对边，如果(a2b2)sin(AB)(a2b2)sin(A B)，判断三角形的形状解：已知等式可化为a2sin(AB)sin(A B)b2 sin(A B)sin(A B)，

22、2a2cosAsinB 2b2cosBsinA.由正弦定理得sin2AcosAsinB sin2BcosBsinA，sinAsinB(sinAcosAsinBcosB)0，sin2A sin2B.由 02A2，02B2 得2A 2B或 2A2B，即 ABC为等腰或直角三角形备选变式（教师专享）已知 ABC中，b2 cosCc2 cosB1cos2C1cos2B，试判断 ABC 的形状解：由已知，得1cos2C1cos2B2cos2C2cos2Bcos2Ccos2Bb2 cosCc2 cosB，cosCcosBbc.由正弦定理知bcsinBsinC，sinBsinCcosCcosB.sinCco

23、sC sinBcosB，即sin2C sin2B，因为 B、C均为 ABC的内角所以2C2B或 2C2B 180，所以 BC或BC 90，故三角形为等腰或直角三角形题型 4 正弦定理、余弦定理的综合应用例 4在ABC中，A、B、C所对的边分别是a、b、c，且 bcosB 是 acosC、ccosA 的等差中项(1)求 B的大小；(2)若 ac10，b2，求 ABC的面积解：(1)由题意，得acosC ccosA 2bcosB.由正弦定理，得sinAcosC cosAsinC 2sinBcosB，即 sin(A C)2sinBcosB.A C B，0B，sin(AC)sinB 0.cosB 12

24、，B 3.(2)由 B3，得a2 c2 b22ac12，即（ac）22ac b22ac12，ac 2.精品w o r d 学习资料可编辑资料-精心整理-欢迎下载-第 4 页，共 8 页文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R

25、6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编

26、码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I

27、3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D

28、9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B

29、10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5

30、C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7

31、L1M10R6A25 SABC12acsinB 32.变式训练已知 a、b、c 分别为 ABC三个内角A、B、C的对边，acosC3asinC bc0.(1)求 A；(2)若 a2，ABC的面积为3，求 b、c.解：(1)由 acosC3asinC bc0 及正弦定理得sinAcosC 3sinAsinC sinBsinC 0.因为 BAC，所以3sinAsinC cosAsinC sinC 0.由于 sinC 0，所以 sinA612.又 0A0)，则b 3t，c 7t，在 ABC 中，由余弦定理得cosCa2 b2c22ab25t29t249t223 5t 3 3t12，所以 C23.2.

32、(20132贵州)ABC的内角 A、B、C 的对边分别为a、b、c，已知 b2，B6，C4，则 ABC的面积为 _答案：31 解析：b 2，B6，C4，由正弦定理得bsin6csin4，解得 c22.又 A(BC)712，SABC12bcsinA 123 23 22362431.3.(20132 盐城期末)在ABC中，若 9cos2A4cos2B5，则BCAC_精品w o r d 学习资料可编辑资料-精心整理-欢迎下载-第 5 页，共 8 页文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L

33、1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6

34、A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码

35、:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3

36、A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9

37、 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B1

38、0U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C

39、2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A26 答案：23解析：由 9cos2A4cos2B5，得 9(1 2sin2A)54(1 2sin2B)，得 9sin2A4sin2B，即 3sinA 2sinB.由正弦定理得BCACsinAsinB23.4.已知 ABC中，B45，AC 4，则 ABC面积的最大值为_答案：442 解析：AC2 AB2BC22AB2BC

40、2 cos45，即 16c2a22ac2 cos45，则有 2ac2ac2 cos45 16，即 ac81cos4516（22）28(2 2)Smax12acsin45 243 8(2 2)442.1.(20142南通一模)在ABC中，a、b、c 分别为角A、B、C所对的边，且 c 3bcosA，tanC34.(1)求 tanB 的值；(2)若 c2，求 ABC的面积解：(1)由正弦定理，得sinC 3sinBcosA，即sin(A B)3sinBcosA.所以sinAcosB cosAsinB 3sinBcosA.从而 sinAcosB 4sinBcosA.因为 cosAcosB0，所以ta

41、nAtanB 4.又 tanC tan(A B)tanA tanBtanAtanB 1，由(1)知，3tanB4tan2B134，解得 tanB12.(2)由(1)，得 sinA 25，sinB 15，sinC 35.由正弦定理，得acsinAsinC232535453.所以 ABC的面积为12acsinB 1234533 231543.2.(20142苏州期末)在ABC中，设角 A、B、C的对边分别为a、b、c，且 acosC12cb.(1)求角 A的大小；(2)若 a15，b4，求边 c 的大小解：(1)用正弦定理，由acosC12cb，精品w o r d 学习资料可编辑资料-精心整理-

42、欢迎下载-第 6 页，共 8 页文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY

43、7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10

44、R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档

45、编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4

46、I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4

47、D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6

48、B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A2文档编码:CK4I3A1S4D9 HT6B10U1M5C2 ZY7L1M10R6A27 得 sinAcosC 12sinC sinB.sinB sin(A C)sinAcosC cosAsinC，12sinC cosAs

49、inC.sinC 0，cosA 12.0A，A3.(2)用余弦定理，得a2b2c22bccosA.a 15，b4，15 16c22343c312.即 c24c10.则 c23.3.在ABC中，A、B、C 所对的边长分别是a、b、c.(1)若 c2，C 3，且 ABC的面积为3，求 a、b 的值；(2)若 sinC sin(B A)sin2A，试判断 ABC 的形状解：(1)c 2，C3，由余弦定理c2a2b22abcosC，得 a2b2ab4.又ABC的面积为3，12absinC 3，即 ab4.联立方程组a2b2ab 4，ab 4，解得 a2，b2.(2)由 sinC sin(B A)sin

50、2A，得 sin(A B)sin(B A)2sinAcosA，即 2sinBcosA2sinAcosA，cosA 2(sinA sinB)0，cosA 0 或 sinA sinB 0.当 cosA0 时，0 A，A2，ABC为直角三角形；当sinA sinB 0 时，得 sinB sinA，由正弦定理得ab，即 ABC为等腰三角形 ABC 为等腰三角形或直角三角形4.在ABC 中，A、B、C所对的边分别为a、b、c，若 a1，b2，cosC14.求：(1)ABC的周长；(2)cos(AC)的值解：(1)因为 c2a2b22abcosC1443144.所以 c2.所以 ABC的周长为 abc 1

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