在锐角三角形中,a b c分别为角A B C所对的边,且根3a=2csinA.确定角C的大小.若c=根3 求三角形ABC

确定角C的大小:
√3a=2c·sinA,由正弦定理 √3sinA = 2sinC·sinA,于是sinC=√3/2.
因为该三角形为锐角三角形,所以C=60°.
若c=√3 求三角形ABC周长的?（最值?题目不全）
c=√3,于是由正弦定理a/sinA = b/sinB = c/sinC = d,
d =√3/(√3/2) = 2
∴a+b+c = d(sinA+sinB+sinC)
= 2(sinA+sinB+√3/2)
= 2[sinA+sin(2π/3-A)+√3/2]
= 2[sinA+sin2π/3·cosA-cos2π/3·sinA+√3/2]
= 2[sinA+√3/2·cosA+1/2sinA+√3/2]
= 2[3/2·sinA+√3/2·cosA+√3/2]
= 2√3sin(A+π/6)+√3
由于A∈(0,2π/3),所以sin(A+π/6)∈(1/2,1]
因此a+b+c∈（2√3,3√3]