cosA=(b^2+c^2-a^2)/2bc
=bc/2bc
=1/2
A=π/3
根據(jù)正弦定理,
b/sinB=a/sinA,a=√3,
A=π/3,B=x,
b/sinx=√3/(√3/2)
b=2sinx,
c/sinC=a/sinA,
c=√3/(√3/2)*sinC
=sinC
=sin(A+B)
=sin(π/3+x)
=√3cosx+sinx
周長(zhǎng):y=a+b+c
=√3+2sinx+√3cosx+sinx
=√3+3sinx+√3cosx
0y=√3+√3(√3sinx+cosx)
=√3+2√3[sinx*cos(π/6)+cosx*sin(π/6)]
=√3+2√3sin(x+π/6)
當(dāng)sin(x+π/6)=1時(shí)函數(shù)有最大值,
y=3√3