作业帮三角形ABC中,A=3/π,BC=3,则三角形ABC周长为?(4根号3)*SIN(B+π/3)+3(4根号3)*SIN(B+π/3)+3
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三角形ABC中,A=3/π,BC=3,则三角形ABC周长为?(4根号3)*SIN(B+π/3)+3(4根号3)*SIN(B+π/3)+3
三角形ABC中,A=3/π,BC=3,则三角形ABC周长为?(4根号3)*SIN(B+π/3)+3
(4根号3)*SIN(B+π/3)+3
三角形ABC中,A=3/π,BC=3,则三角形ABC周长为?(4根号3)*SIN(B+π/3)+3(4根号3)*SIN(B+π/3)+3
A=π/3 C=π-A-B=2π/3-B
由正弦定理 AC/sinB=AB/sinC=BC/sinA=3/(√3/2)=2√3
所以AC=2√3sinB AB=2√3sinC=2√3sin(2π/3-B)
AB+AC=2√3*2sin(π/3)cos(π/3-B)=6sin[π/2-(π/3-B)]=6sin(B+π/6)
所以周长=AB+AC+BC=6sin(B+π/6)+3
我觉得你给的答案错了
A=π/3，BC=3，
根据正弦定理：a/sinA=b/sinB=c/sinC,
所以3/(sinπ/3)= b/sinB=c/sin(A+B),
3/(sinπ/3)= b/sinB=c/sin(π/3+B),
2√3= b/sinB=c/sin(π/3+B),
所以b=2√3 sinB，c=2√3 sin(π/3+B),
∴三角形ABC周长=a+...
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A=π/3，BC=3，
根据正弦定理：a/sinA=b/sinB=c/sinC,
所以3/(sinπ/3)= b/sinB=c/sin(A+B),
3/(sinπ/3)= b/sinB=c/sin(π/3+B),
2√3= b/sinB=c/sin(π/3+B),
所以b=2√3 sinB，c=2√3 sin(π/3+B),
∴三角形ABC周长=a+b+c
=3+2√3 sinB+2√3 sin(π/3+B)
=3+2√3 sinB+2√3(sinπ/3cos B+ cosπ/3sin B)
=3+3√3 sinB+3 cos B
=3+6(√3/2 sinB+1/2 cos B)
=3+6 sin(B+π/6).
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