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已知锐角三角形ABC中,sin(A+B)=3/5,sin (A%B)=1/5,tAnA=2tAnB,...

sin(a b) sin(a-b)=2sinacosb=3/5 1/5=4/5 所以sinacosb=2/5 sin(a b)-sin(a-b)=2cosasinb=3/5-1/5=2/5 所以cosasinb=1/5 (sinacosb)/(cosasinb)=(2/5)/(1/5)=2 即tgactgb=2 设ab边上的高为cd,d为垂足 则tgactgb=(cd/ad)*(bd/cd)=bd/ad=2 因为ab=3 所

(I)证明:∵sin(A+B)= 3 5 ,sin(A-B)= 1 5 ,∴sinAcosB+cosAsinB= 3 5 ,sinAcosB-cosAsinB= 1 5 ,∴sinAcosB= 2 5 ,cosAsinB= 1 5 ,∴tanA=2tanB.(2)∵ π 2 3 5 ,∴ cos(A+B)=- 4 5 , tan(A+B)=- 3 4 即 tanA+tanB 1-tanAtanB =- 3 4 ,将

(1)sin(A+B)=3/5,sin(A-B)=1/5sin(a+b)=sinAcosB+sinBcosA=3/5sin(a-b)=sinAcosB-sinBcosA=1/5两式相加相减后可得:sinAcosB=2/5sinBcosA=1/5将两式相除,可得tanA=2tanB(2)tan(B)=sinB/cosB=sinBcosA/cosAcosBcos(A)cos(B)=1/2[cos(A+B)+cos(A-B)]=1/2[4/5+2根号6/5]=(根号6-2)/5tanB=1/(根号6-2)=(根号6+2)/2

解:sin(A+B)=sinAcosB+sinBcosA=3/5(1) sin(A-B)=sinAcosB-sinBcosA=1/5(2) (1)=3*(2) sinAcosB+sinBcosA=3sinAcosB-3sinBcosA 2sinBcosA=sinAcosB tanA/tanB =2.望采纳

sin(A+B)=3/5,sin(A-B)=1/5则:sin(A+B)=3sin(A-B)sinAcosB+cosAsinB=3sinAcosB-3cosAsinB2sinAcosB=4cosAsinBsinA/cosA=2sinB/cosBtanA=2tanB(2)sin(A+B)=3/5sinC=sin[180°-(A+B)]=sin(A+B)=3/5则:cosC=根号(1-sinC)=4/5则:

sin(A B) sin(A-B)=2sinAcosB=3/5 1/5=4/5 所以sinAcosB=2/5 sin(A B)-sin(A-B)=2cosAsinB=3/5-1/5=2/5 所以cosAsinB=1/5 (sinAcosB)/(cosAsinB)=(2/5)/(1/5)=2 即tgActgB=2 设AB边上的高为CD,D为垂足 则tgActgB=(CD/AD)*(BD/CD)=BD/AD=

(1)sin(A+B)=3/5,sin(A-B)=1/5sin(a+b)=sinAcosB+sinBcosA=3/5sin(a-b)=sinAcosB-sinBcosA=1/5两式相加相减后可得:sinAcosB=2/5sinBcosA=1/5将两式相除,可得tanA=2tanB(2)tan(B)=sinB/cosB=sinBcosA/cosAcosBcos(A)cos(B)=1/2[cos(A+B)+cos(A-B)]=1/2[4/5+2根号6/5]=(根号6-2)/5tanB=1/(根号6-2)=(根号6+2)/2

(I)∵ sin(A+B)=sinAcosB+cosAsinB= 3 5 ① sin(A-B)=sinAcosB-cosAsinB= 1 5 ②①+②得: 2sinAcosB= 4 5 ,∴ sinAcosB= 2 5 ③ cosAsinB= 1 5 ④③/④得:tanA?cotB=2,即 tanA tanB =2 (II)∵△ABC是锐角三角形,又

证明:由已知得sin(A+B)=sinA*cosB+sinB*cosA=3/5 (1)sin(A-B)=sinA*cosB-sinB*cosA=1/5 (2)(1)+(2)得sinA*cosB=2/5 (1)-(2)得sinB*cosA=1/5 tanA/tanB=(sinA*cosB)/(sinB*cosA)=(2/5)/(1/5)=2所以tanA=2tanB

sin(A+B)=sinAcosB+cosAsinB=3/5(1)sin(A-B)=sinAcosB-cosAsinB=1/5(2)(1)-(2)*3可得2sinAcosB=4osAsinB,两边同时除以cosBcosA所以tan A=2tanB作AB边上的高CD交AB于D,设CD=h,AD=X,BD=3-Xtan A=2tanB ,

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