设A、B、C 为三角形的内角,且方程(sinB-sinA)x^2+(sinA-sinC)x^2+(sinC-sinB)=0有等根,那么角B( )A.B>60° B.B≥60° C.B<60°D.B≤60°方程应该是(sinB-sinA)x^2+(sinA-sinC)x+(sinC-sinB)=0不好意思…
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/02 20:01:41
![设A、B、C 为三角形的内角,且方程(sinB-sinA)x^2+(sinA-sinC)x^2+(sinC-sinB)=0有等根,那么角B( )A.B>60° B.B≥60° C.B<60°D.B≤60°方程应该是(sinB-sinA)x^2+(sinA-sinC)x+(sinC-sinB)=0不好意思…](/uploads/image/z/5394540-12-0.jpg?t=%E8%AE%BEA%E3%80%81B%E3%80%81C+%E4%B8%BA%E4%B8%89%E8%A7%92%E5%BD%A2%E7%9A%84%E5%86%85%E8%A7%92%2C%E4%B8%94%E6%96%B9%E7%A8%8B%EF%BC%88sinB-sinA%29x%5E2%2B%28sinA-sinC%29x%5E2%2B%28sinC-sinB%29%3D0%E6%9C%89%E7%AD%89%E6%A0%B9%2C%E9%82%A3%E4%B9%88%E8%A7%92B%28+%29A.B%EF%BC%9E60%C2%B0+B.B%E2%89%A560%C2%B0+C.B%EF%BC%9C60%C2%B0D.B%E2%89%A460%C2%B0%E6%96%B9%E7%A8%8B%E5%BA%94%E8%AF%A5%E6%98%AF%28sinB-sinA%29x%5E2%2B%28sinA-sinC%29x%2B%28sinC-sinB%29%3D0%E4%B8%8D%E5%A5%BD%E6%84%8F%E6%80%9D%E2%80%A6)
设A、B、C 为三角形的内角,且方程(sinB-sinA)x^2+(sinA-sinC)x^2+(sinC-sinB)=0有等根,那么角B( )A.B>60° B.B≥60° C.B<60°D.B≤60°方程应该是(sinB-sinA)x^2+(sinA-sinC)x+(sinC-sinB)=0不好意思…
设A、B、C 为三角形的内角,且方程(sinB-sinA)x^2+(sinA-sinC)x^2+(sinC-sinB)=0有等根,那么角B( )
A.B>60°
B.B≥60°
C.B<60°
D.B≤60°
方程应该是(sinB-sinA)x^2+(sinA-sinC)x+(sinC-sinB)=0不好意思…
设A、B、C 为三角形的内角,且方程(sinB-sinA)x^2+(sinA-sinC)x^2+(sinC-sinB)=0有等根,那么角B( )A.B>60° B.B≥60° C.B<60°D.B≤60°方程应该是(sinB-sinA)x^2+(sinA-sinC)x+(sinC-sinB)=0不好意思…
D
方程应该是(sinB-sinA)x^2+(sinA-sinC)x+(sinC-sinB)=0吧?
∵方程有等根
∴△=0
即(sinA-sinc)^2-4(sinB-sinA)(sinC-sinB)=0 ①
由正弦定理a=2RsinA,得
sinA=a/(2R)
sinB=b/(2R)
sinC=c/(2R)
将上面3个式子带入①,...
全部展开
方程应该是(sinB-sinA)x^2+(sinA-sinC)x+(sinC-sinB)=0吧?
∵方程有等根
∴△=0
即(sinA-sinc)^2-4(sinB-sinA)(sinC-sinB)=0 ①
由正弦定理a=2RsinA,得
sinA=a/(2R)
sinB=b/(2R)
sinC=c/(2R)
将上面3个式子带入①,得
(a-c)^2-4(b-a)(c-b)=0
a^2+2ac+c^2+4b^2-4ab-4bc=0
(a+c)^2-4b(a+c)+4b^2=0
(a+c-2b)^2=0
a+c=2b
cosB
=(a^2+c^2-b^2)/(2ac)
={a^2+c^2-[(a+c)/2]^2}/(2ac)
=3(a^2+c^2)/(8ac)-1/4
=(3×2ac)/(8ac)-1/4
即cosB≥1/2
所以B的范围是(0,π/3]
即0<B≤60°
没有正确选项
收起
你这个方程不对吧……