在△ABC ,角ABC对应边分别是abc,且c=2 C=60°求(a+b)/(sinA+sinB)
1个回答
展开全部
(1)sina+siinb=2sin(a+b)
==>sina+√3/2=2sin(a+π/3)
==>sina+√3/2=sina+√3cosa
==>cosa=1/2
则a=π/3
则c=π-a-b=π/3
(2)由题意可得a+b=2c
==>c=(a+b)/2
cosc=(a²+b²-c²)/2ab
=[a²+b²-((a+b)/2)²]/2ab
=3a/8b+3b/8a-1/4
≥2√(3a/8b)*(3b/8a)-1/4
=1/2
所以c∈(0,π/3]
则c的最大值π/3
==>sina+√3/2=2sin(a+π/3)
==>sina+√3/2=sina+√3cosa
==>cosa=1/2
则a=π/3
则c=π-a-b=π/3
(2)由题意可得a+b=2c
==>c=(a+b)/2
cosc=(a²+b²-c²)/2ab
=[a²+b²-((a+b)/2)²]/2ab
=3a/8b+3b/8a-1/4
≥2√(3a/8b)*(3b/8a)-1/4
=1/2
所以c∈(0,π/3]
则c的最大值π/3
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
