在△ABC中,内角A.B.C所对的边为a.b.c。tanC=3/4,c=-3bcosA
在△ABC中,内角A.B.C所对的边为a.b.c。tanC=3/4,c=-3bcosA求tanB若c=2,求△ABC面积...
在△ABC中,内角A.B.C所对的边为a.b.c。tanC=3/4,c=-3bcosA求tanB
若c=2,求△ABC面积 展开
若c=2,求△ABC面积 展开
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1)c=-3bcosA
sinC=-3sinBcosA
sin(A+B)=-3sinBcosA
sinAcosB+sinBcosA=-3sinBcosA
sinAcosB=-4sinBcosA
tanA=-4tanB
tanC=-tan(A+B)=-(tanA+tanB)/(1-tanAtanB)=3/4
-(-4tanB+tanB)/[1+4(tanB)^2]=3/4
tanB/[1+4(tanB)^2]=1/4
4(tanB)^2-4tanB+1=0
(2tanB-1)^2=0
tanB=1/2
2)c=-3bcosA
A是钝角,
tanC=sinC/cosC=3/4 ,
cosC=4sinC/3 ,
(cosC)^2+(sinC)^2=1
16(sinC)^2/9+(sinC)^2=1
sinC=3/5 ,cosC=4/5
tanB=sinB/cosB=1/2 ,
cosB=2sinB,
(cosB)^2+(sinB)^2=1
4(sinB)^2+(sinB)^2=1
sinB=√5/5 ,cosB=2√5/5
sinA=sin(B+C)=sinBcosC+cosBsinC=√5/5*4/5+2√5/5*3/5=2√5/2
2R=b/sinB=a/sinA=c/sinC=2/(3/5)=10/3
a=2RsinA,b=2RsinB
S=1/2*absinC=1/2*(2R)^2*sinAsinBsinC
=1/2*(10/3)^2*2√5/2*√5/5 *3/5
=8/3
sinC=-3sinBcosA
sin(A+B)=-3sinBcosA
sinAcosB+sinBcosA=-3sinBcosA
sinAcosB=-4sinBcosA
tanA=-4tanB
tanC=-tan(A+B)=-(tanA+tanB)/(1-tanAtanB)=3/4
-(-4tanB+tanB)/[1+4(tanB)^2]=3/4
tanB/[1+4(tanB)^2]=1/4
4(tanB)^2-4tanB+1=0
(2tanB-1)^2=0
tanB=1/2
2)c=-3bcosA
A是钝角,
tanC=sinC/cosC=3/4 ,
cosC=4sinC/3 ,
(cosC)^2+(sinC)^2=1
16(sinC)^2/9+(sinC)^2=1
sinC=3/5 ,cosC=4/5
tanB=sinB/cosB=1/2 ,
cosB=2sinB,
(cosB)^2+(sinB)^2=1
4(sinB)^2+(sinB)^2=1
sinB=√5/5 ,cosB=2√5/5
sinA=sin(B+C)=sinBcosC+cosBsinC=√5/5*4/5+2√5/5*3/5=2√5/2
2R=b/sinB=a/sinA=c/sinC=2/(3/5)=10/3
a=2RsinA,b=2RsinB
S=1/2*absinC=1/2*(2R)^2*sinAsinBsinC
=1/2*(10/3)^2*2√5/2*√5/5 *3/5
=8/3
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