"在△ABC中,内角A,B,C的对边分别为A,B,C已知CosA=2/3,SinB=√5cosC,求TanC"
2个回答
展开全部
(1)
COSA = 2/3,新浪=√5/3,塔纳=√5/2
SINB = SIN(π-AC)= SIN(A + C)= sinAcosC + cosAsinC = √5/3cosC +2 / 3sinC
由SINB =√5cosC称为
所以√5cosC =√5/3cosC +2 / 3sinC
另外COSC得到等号两边
>√5 =√5/3 +2 / 3tanC
TANC =√5
(2)在B中的AC垂直,踏板到D
TANC =√5,正弦=√5 /√6,COSC = 1 /√6
然后是CD = BC * COSC = 1 /√3,BD = BC * SINC =√5 /√3
AD = BD /塔纳= 2 /√3
AC = AD + CD =√3
的面积三角形ABC = 1/2 * AC * BD = 1/2 *√3 *√5 /√3 =√5 / 2
COSA = 2/3,新浪=√5/3,塔纳=√5/2
SINB = SIN(π-AC)= SIN(A + C)= sinAcosC + cosAsinC = √5/3cosC +2 / 3sinC
由SINB =√5cosC称为
所以√5cosC =√5/3cosC +2 / 3sinC
另外COSC得到等号两边
>√5 =√5/3 +2 / 3tanC
TANC =√5
(2)在B中的AC垂直,踏板到D
TANC =√5,正弦=√5 /√6,COSC = 1 /√6
然后是CD = BC * COSC = 1 /√3,BD = BC * SINC =√5 /√3
AD = BD /塔纳= 2 /√3
AC = AD + CD =√3
的面积三角形ABC = 1/2 * AC * BD = 1/2 *√3 *√5 /√3 =√5 / 2
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询