已知三角形ABC中,A.B.C的对边分别是a.b.c,b+c=√3a 设向量m=(cos(派/2+A),-1),向量n=(cosA-5/4,-sinA已知三角形ABC中,A.B.C的对边分别是a.b.b+c=√3a 设向量m=(cos(派/2+A),-1),向量n=(cosA-5/4,-sinA)向量m//向量n
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![已知三角形ABC中,A.B.C的对边分别是a.b.c,b+c=√3a 设向量m=(cos(派/2+A),-1),向量n=(cosA-5/4,-sinA已知三角形ABC中,A.B.C的对边分别是a.b.b+c=√3a 设向量m=(cos(派/2+A),-1),向量n=(cosA-5/4,-sinA)向量m//向量n](/uploads/image/z/3716994-66-4.jpg?t=%E5%B7%B2%E7%9F%A5%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2CA.B.C%E7%9A%84%E5%AF%B9%E8%BE%B9%E5%88%86%E5%88%AB%E6%98%AFa.b.c%2Cb%2Bc%3D%E2%88%9A3a+%E8%AE%BE%E5%90%91%E9%87%8Fm%3D%28cos%28%E6%B4%BE%2F2%2BA%29%2C-1%29%2C%E5%90%91%E9%87%8Fn%3D%EF%BC%88cosA-5%2F4%2C-sinA%E5%B7%B2%E7%9F%A5%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%EF%BC%8CA.B.C%E7%9A%84%E5%AF%B9%E8%BE%B9%E5%88%86%E5%88%AB%E6%98%AFa.b.b%2Bc%3D%E2%88%9A3a+%E8%AE%BE%E5%90%91%E9%87%8Fm%3D%28cos%28%E6%B4%BE%2F2%2BA%29%2C-1%29%2C%E5%90%91%E9%87%8Fn%3D%EF%BC%88cosA-5%2F4%2C-sinA%EF%BC%89%E5%90%91%E9%87%8Fm%2F%2F%E5%90%91%E9%87%8Fn)
已知三角形ABC中,A.B.C的对边分别是a.b.c,b+c=√3a 设向量m=(cos(派/2+A),-1),向量n=(cosA-5/4,-sinA已知三角形ABC中,A.B.C的对边分别是a.b.b+c=√3a 设向量m=(cos(派/2+A),-1),向量n=(cosA-5/4,-sinA)向量m//向量n
已知三角形ABC中,A.B.C的对边分别是a.b.c,b+c=√3a 设向量m=(cos(派/2+A),-1),向量n=(cosA-5/4,-sinA
已知三角形ABC中,A.B.C的对边分别是a.b.b+c=√3a 设向量m=(cos(派/2+A),-1),向量n=(cosA-5/4,-sinA)向量m//向量n
求 A B C
已知三角形ABC中,A.B.C的对边分别是a.b.c,b+c=√3a 设向量m=(cos(派/2+A),-1),向量n=(cosA-5/4,-sinA已知三角形ABC中,A.B.C的对边分别是a.b.b+c=√3a 设向量m=(cos(派/2+A),-1),向量n=(cosA-5/4,-sinA)向量m//向量n
∵向量m‖向量n,
∴cos(π/2+A)*(-sinA)=-1*(cosA-5/4)
sin(-A)*(-sinA)=5/4-cosA
sin²A=5/4-cosA
1-cos²A=5/4-cosA
cos²A-cosA+1/4=0
(cosA-1/2)²=0
∴cosA=1/2
A=π/3
sinB+sinC=√3*sinA=3/2
B+C=2π/3……①
sinB+sinC=2sin[(B+C)/2]cos[(B-C)/2]=3/2
∴cos[(B-C)/2]=√3/2
B和C是对称的不妨设B>C
∴(B-C)/2=π/6
B-C=π/3……②
由①和②,得
B=π/2,C=π/6
即A=A=π/3,B=π/2,C=π/6
谢谢