在△ABC中,A=π/3,BC=3,则三角形周长为S 证明：S=6sin（B+π/6）+3
问题描述：
在△ABC中,A=π/3,BC=3,则三角形周长为S 证明：S=6sin（B+π/6）+3
答
过点C作CD垂直AB交AB于D，
BD=3cosB CD=3sinB AD=CDcotA =3sinB*cot(π/3) AC=2AD
S=AD+DB+AC+BC
=3sinB*cot(π/3)+3cosB+6sinB*cot(π/3)+3
=3(cosB+根号3*sinB)+3
=6(1/2cosB+根号3/2*sinB)+3
=6(sinπ/6)cosB+cosπ/6sinB)+3
=6sin(B+π/6)+3
答
由正弦定理：a/sinA=b/sinB=c/sinC
∵BC=a=3,A=π/3
∴a/sinA=b/sinB=c/sinC=3/√3/2=2√3
∴b=2√3sinB,c=2√3sinC
∵A+B+C=π,A=π/3
∴C=2π/3-B
∴c=2√3sin(2π/3-B)
三角形ABC周长=a+b+c
=3+2√3sinB+2√3sin(2π/3-B)
=3+6sin(B+π/6)