tan(π/4+B)=[tan(π/4)+tanB]/[1-tan(π/4)tanB]=-3
得:
tanB=2
因为:A=π/3、tanB=2
则:
tanC
=tan[π-(A+B)]
=-tan(A+B)
=-tan(π/3+B)
=-[tan(π/3)+tanB]/[1-tan(π/3)tanB]
=-[√3+2]/[1-2√3]
=(8+5√3)/11
tan(π/4+B)=[tan(π/4)+tanB]/[1-tan(π/4)tanB]=-3
得:
tanB=2
因为:A=π/3、tanB=2
则:
tanC
=tan[π-(A+B)]
=-tan(A+B)
=-tan(π/3+B)
=-[tan(π/3)+tanB]/[1-tan(π/3)tanB]
=-[√3+2]/[1-2√3]
=(8+5√3)/11