作业帮在三角形ABC中A=TT/3,BC=3,则AB+AC的长可表示为
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在三角形ABC中A=TT/3,BC=3,则AB+AC的长可表示为
在三角形ABC中A=TT/3,BC=3,则AB+AC的长可表示为
在三角形ABC中A=TT/3,BC=3,则AB+AC的长可表示为
a/sinA=b/sinB=c/sinC b=2根号3*sinB c=2根号3*sinC
b+c=2根号3(sinB+sin(120度-B))=2根号3(sinB+根号3/2cosB+1/2sinB)
=2根号3(3/2sinB+根号3/2cosB)=6Sin(B+30度)
=6Sin(B+π/6)
TT/3是什么
由正弦定理，得
3/sin(π/3)=AB/sinC=AC/sinB
又由等比定理，得
3/sin(π/3)=AB/sinC=AC/sinB=(AB+AC)/(sinB+sinC)
∴AB+AC=2√3(sinB+sinC)=2√3[sinB+sin(2π/3-B)]
=2√3[sinB+sin2π/3cosB-cos2π/3sinB]
=2√3[s...
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由正弦定理，得
3/sin(π/3)=AB/sinC=AC/sinB
又由等比定理，得
3/sin(π/3)=AB/sinC=AC/sinB=(AB+AC)/(sinB+sinC)
∴AB+AC=2√3(sinB+sinC)=2√3[sinB+sin(2π/3-B)]
=2√3[sinB+sin2π/3cosB-cos2π/3sinB]
=2√3[sinB+√3/2cosB+sinB/2]
=2√3[3sinB/2+√3/2cosB]
=6(√3/2sinB+cosB/2)
=6(sinBcosπ/6+cosBsinπ/6)
=6sin(B+π/6)
收起
正弦定理
AB/sinC=BC/sinA AB=2√3*sinC
同理
AC/sinB=BC/sinA AC=2√3*sinB
AB+AC=2√3(sinC+sinB) A=π/3 C=2π/3-B
=2√3(sin(2π/3-B)+sinB)
=2√3(√3/2*cosB-1/2sinB+sinB)
=2√3(√3/2*cosB+3/2sinB)
=6sin(B+π/6)