如题所述
(1)
(b+c)²+(-1)(a²+bc)=0
b²+2bc+c²-a²-bc=0
b²+c²-a²=-bc
由余弦定理得：
cosA=(b²+c²-a²)/(2bc)=-bc/(2bc)=-½
A为三角形内角，A=2π/3
(2)
由余弦定理得：
cosA=(b²+c²-a²)/(2bc)=[(b+c)²-2bc-a²]/(2bc)
a=3，cosA=-½代入，得：
[(b+c)²-2bc-3²]/(2bc)=-½
整理，得：bc=(b+c)²-9
由均值不等式得：bc≤(b+c)²/4
(b+c)²-9≤(b+c)²/4
整理，得：(b+c)²≤12
b+c≤2√3
a+b+c≤3+2√3
△ABC周长的最大值为3+2√3
温馨提示：答案为网友推荐，仅供参考
第1个回答 2016-09-02
2)
A=2π/3,B+C=π/3,
2R=a/sinA=b/sinB=c/sinC
2R=a/sinA=3/sin2π/3=3/√3/2=2√3
a=3 ,b=2RsinB ,c=2RsinC
a+b+c=a+2RsinB+2RsinC
=a+2R(sinB+sinC)
=a+2R[sinB+sin(π/3-B)]
=3+2√3[sinB+√3/2cosB-1/2sinB]
=3+2√3(1/2sinB+√3/2cosB)
=3+2√3sin(B+π/3)
0<B<π/3
π/3<B+π/3<2π/3
B+π/3∈(π/3，2π/3)
sin(B+π/3)在B+π/3∈(π/3，2π/3)的值域为:(√3/2,1]
所以,3+2√3sin(B+π/3)的值域为:(3,3+2√3]
即周长最大值为:3+2√3
A=2π/3,B+C=π/3,
2R=a/sinA=b/sinB=c/sinC
2R=a/sinA=3/sin2π/3=3/√3/2=2√3
a=3 ,b=2RsinB ,c=2RsinC
a+b+c=a+2RsinB+2RsinC
=a+2R(sinB+sinC)
=a+2R[sinB+sin(π/3-B)]
=3+2√3[sinB+√3/2cosB-1/2sinB]
=3+2√3(1/2sinB+√3/2cosB)
=3+2√3sin(B+π/3)
0<B<π/3
π/3<B+π/3<2π/3
B+π/3∈(π/3，2π/3)
sin(B+π/3)在B+π/3∈(π/3，2π/3)的值域为:(√3/2,1]
所以,3+2√3sin(B+π/3)的值域为:(3,3+2√3]
即周长最大值为:3+2√3
第2个回答 2016-09-02
等腰三角形，2倍根号3+3
