高中数学题！求解
1个回答
展开全部
(1)
a/sinA =b/sin2B
a/sinA =b/(2sinB.cosB)
1/(2cosB) = 1
cosB =1/2
B=π/3
(2)
b=√13
a-c=2
(a-c)^2 =4
a^2+c^2 -2ac =4
[a^2+c^2 -2ac.cos(π/3)] -2ac.cos(π/3)= 4
b^2 -2ac.cos(π/3)=4
13- ac =4
ac = 9
SΔABC
=(1/2)ac.sinB
=(1/2)(9)(√3/2)
=9√3/4
a/sinA =b/sin2B
a/sinA =b/(2sinB.cosB)
1/(2cosB) = 1
cosB =1/2
B=π/3
(2)
b=√13
a-c=2
(a-c)^2 =4
a^2+c^2 -2ac =4
[a^2+c^2 -2ac.cos(π/3)] -2ac.cos(π/3)= 4
b^2 -2ac.cos(π/3)=4
13- ac =4
ac = 9
SΔABC
=(1/2)ac.sinB
=(1/2)(9)(√3/2)
=9√3/4
本回答被提问者和网友采纳
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
