已经cosB=3/5,得sinB=√(1-cosB^2)=2/5
从而tanB=sinB/cosB=2/3
tab(A-B)=(tanA-tanB)/(1+tanAtanB)
即(tanA-2/3)/(1+2/3tabA)=-1/3
tanA=3/11
tanC=tan(180-A-B)=-tan(A+B)=-(tanA+tanB)/(1-tanAtanB)=-(3/11+2/3)/(1-3/11X2/3)=-31/27
已经cosB=3/5,得sinB=√(1-cosB^2)=2/5
从而tanB=sinB/cosB=2/3
tab(A-B)=(tanA-tanB)/(1+tanAtanB)
即(tanA-2/3)/(1+2/3tabA)=-1/3
tanA=3/11
tanC=tan(180-A-B)=-tan(A+B)=-(tanA+tanB)/(1-tanAtanB)=-(3/11+2/3)/(1-3/11X2/3)=-31/27