问题
填空题
已知a,b,c分别为△ABC三个内角A,B,C所对的边长,且acosB-bcosA=
|
答案
∵△ABC中acosB-bcosA=
c,2 5
∴根据正弦定理,得sinAcosB-sinBcosA=
sinC2 5
∵sinC=sin(A+B)=sinAcosB+sinBcosA
∴sinAcosB-sinBcosA=
(sinAcosB+sinBcosA),解之得3sinAcosB=7sinBcosA2 5
因此,
=tanA tanB
=sinA cosA sinB cosB
=sinAcosB cosAsinB
=7sinAcosB 7cosAsinB
=7sinAcosB 3sinAcosB 7 3
故答案为:7 3