已知在三角形ABC中,abc分别为角ABC的对边,且2【1-cos(B+C)】-cos2A=7/2(1)若sinA=2sinBcosC,试判断三角形ABC的形状(2)若a=根号3,b+c=3,求b和c的值
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/22 16:59:20
![已知在三角形ABC中,abc分别为角ABC的对边,且2【1-cos(B+C)】-cos2A=7/2(1)若sinA=2sinBcosC,试判断三角形ABC的形状(2)若a=根号3,b+c=3,求b和c的值](/uploads/image/z/2583651-3-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2Cabc%E5%88%86%E5%88%AB%E4%B8%BA%E8%A7%92ABC%E7%9A%84%E5%AF%B9%E8%BE%B9%2C%E4%B8%942%E3%80%901-cos%EF%BC%88B%2BC%EF%BC%89%E3%80%91-cos2A%3D7%2F2%EF%BC%881%EF%BC%89%E8%8B%A5sinA%3D2sinBcosC%2C%E8%AF%95%E5%88%A4%E6%96%AD%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E5%BD%A2%E7%8A%B6%EF%BC%882%EF%BC%89%E8%8B%A5a%3D%E6%A0%B9%E5%8F%B73%2Cb%2Bc%3D3%2C%E6%B1%82b%E5%92%8Cc%E7%9A%84%E5%80%BC)
已知在三角形ABC中,abc分别为角ABC的对边,且2【1-cos(B+C)】-cos2A=7/2(1)若sinA=2sinBcosC,试判断三角形ABC的形状(2)若a=根号3,b+c=3,求b和c的值
已知在三角形ABC中,abc分别为角ABC的对边,且2【1-cos(B+C)】-cos2A=7/2
(1)若sinA=2sinBcosC,试判断三角形ABC的形状
(2)若a=根号3,b+c=3,求b和c的值
已知在三角形ABC中,abc分别为角ABC的对边,且2【1-cos(B+C)】-cos2A=7/2(1)若sinA=2sinBcosC,试判断三角形ABC的形状(2)若a=根号3,b+c=3,求b和c的值
先解出cosA
因为cos(B+C) = cos(π -A) = - cosA
cos2A = 2cos²A -1
解方程
2(1+cosA) -(2cos²A-1) = 7/2
化简即 2cos²A - 2cosA + 1/2 =0
cosA = 1/2
(1)
2sinBcosC = sin(B+C) - sin(C-B) = sinA - sin(C-B)
所以 sinA - sin(C-B) = sinA
sin(C-B) =0
C=B
又因为A=π/3 , 所以C=B= (π-π/3)/2=π/3
ABC是等边三角形
(2)
余弦定理\
a² = b² +c² -2bc cosA = b²+c² -bc
所以 3 = b²+c² -bc
b+c=3, 所以(b+c)² = b²+c² +2bc =9
两式相减
3bc = 9-3 =6
bc =2
故bc =2,
b+c =3
解之得
b=1, c=2或
b=2, c=1