www.fltk.net > 在△ABC中,内角A,B,C的对边分别为A,B,C已知CosA=2分之3,sinB=根号5CosC

在△ABC中,内角A,B,C的对边分别为A,B,C已知CosA=2分之3,sinB=根号5CosC

)cosA=2/3,sinA=√5/3,tanA=√5/2sinB=sin(π-A-C)=sin(A+C)=sinAcosC+cosAsinC=√5/3cosC+2/3sinC由已知sinB=√5cosC所以√5cosC=√5/3cosC+2/3sinC等号两边同时除cosC得√5=√5/3+2/3tanCtanC=√5 (2)过B作AC垂线,垂足为DtanC=√5,sinC=√5/√6, cosC=1/√6则CD=BC*cosC=1/√3,BD=BC*sinC=√5/√3AD=BD/tanA=2/√3AC=AD+CD=√3三角形ABC的面积=1/2*AC*BD=1/2*√3*√5/√3=√5/2

(1) ∵cosA=2/3,∴sinA=√(1-cosA)=√5/3 ∵sinB=√5cosC sinB=sin(A+C)=sinAcosC+cosAsinC∴sinAcosC+cosAsinC=√5cosC ∴√5/3cosC+2/3sinC=√5cosC∴ sinC=√5cosC ,∴tanC=√5(2)若a=√2,∵ sinA=√5/3∴2R=a/sinA=√2/(√

1∵cosa=2/3,∴sina=√(1-cosa)=√5/3∵sinb=√5cosc sinb=sin(a+c)=sinacosc+cosasinc∴sinacosc+cosasinc=√5cosc ∴√5/3cosc+2/3sinc=√5cosc∴ sinc=√5cosc ,∴tanc=√52.若a=√2,∵ sina=√5/3∴2r=a/sina=√2/(√5/3)=3√10/5∵

(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

∵cosA=2/3∴sinA=√(1-cosA) = √5/3∵sinB=√5cosC又,sinB=sin(A+C)=sinAcosC+cosAsinC∴√5cosC=√5/3cosC+2/3sinC∴3√5cosC=√5cosC+2sinC∴2√5cosC=2sinC∴tanC=√5

(1)∵A为三角形的内角,cosA=2 3 ,∴sinA= 1?cos2A = 5 3 ,又 5 cosC=sinB=sin(A+C)=sinAcosC+cosAsinC= 5 3 cosC+2 3 sinC,整理得:2 5 3 cosC=2 3 sinC,则tanC= 5 ;(2)由tanC= 5 得:cosC= 1 sec2C = 1 1+tan2C = 1 1+5 = 6 6 ,∴

(1)因为(cosa-2cosc)/cosb=(2c-a)/b所以(cosa-2cosc)/cosb=(2sinc-sina)/sinb cosasinb-2sinbcosc=2sinccosb-sinacosb cosasinb+sinacosb=2(sinbcosc+sinccosb) sinc=2sina 所以sinc/sina=2(2)c/a=sinc/sina=2 c=2a b=2 cosb=(a^2+c^2-b^2)/(2

解答:(cosA-2cosC)/cosB=(2c-a)/b根据正弦定理(cosA-2cosC)/cosB=(2sinC-sinA)/sinB∴sinBcosA-2cosCsinB=2sinCcosB-sinAcosB∴sinBcosA+cosBsinA=2(sinBcosC+cosBsinC)∴sin(B+A)=2sin(B+C)∴sinC=2sinA∴sinC/sinA=2

解:∵cosA=-√2/4 ∴sinA=√14/4 由正弦定理,有 a/sinA=c/sinC 则 sinC=c*sinA/a =√2*(√14/4)÷2 =√7/4 cosC=3/4 ∵sinB=sin[π-(A+C)]=sin(A+C) ∴sinB=sinA*cosC+cosA*sinC =(√14/4)*(3/4)+(-√2/4)*(√7/4) =√14/8 故 b=a*sinB/sinA =2*

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