锐角三角形ABC中,a,b,c为角ABC所对的边,且(b-2c)cosA=a-2acos^2(B/2)求角A.若a=根号3,求b+c的取值范围
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![锐角三角形ABC中,a,b,c为角ABC所对的边,且(b-2c)cosA=a-2acos^2(B/2)求角A.若a=根号3,求b+c的取值范围](/uploads/image/z/1600520-32-0.jpg?t=%E9%94%90%E8%A7%92%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2Ca%2Cb%2Cc%E4%B8%BA%E8%A7%92ABC%E6%89%80%E5%AF%B9%E7%9A%84%E8%BE%B9%2C%E4%B8%94%28b-2c%29cosA%3Da-2acos%5E2%28B%2F2%29%E6%B1%82%E8%A7%92A.%E8%8B%A5a%3D%E6%A0%B9%E5%8F%B73%2C%E6%B1%82b%2Bc%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
锐角三角形ABC中,a,b,c为角ABC所对的边,且(b-2c)cosA=a-2acos^2(B/2)求角A.若a=根号3,求b+c的取值范围
锐角三角形ABC中,a,b,c为角ABC所对的边,且(b-2c)cosA=a-2acos^2(B/2)
求角A.若a=根号3,求b+c的取值范围
锐角三角形ABC中,a,b,c为角ABC所对的边,且(b-2c)cosA=a-2acos^2(B/2)求角A.若a=根号3,求b+c的取值范围
(b-2c)cosA=a-2acos^2(B/2)
则(sinB-2sinC)cosA=sinA-sinA(1+cosB)
则sinBcosA-2sinCcosA=sinA-sinA-sinAcosB
sinBcosA+cosBsinA-2sinCcosA=0
sin(B+A)=2sinCcosA
sinC=2sinCcosA
cosA=1/2
A=60°
a=√3
由余弦定理得a^2=b^2+c^2-2bccosA
=b^2+c^2-bc>=2bc-bc=bc
即bc<=a^2=3
而(b+c)^2=b^2+c^2+2bc=a^2+3bc<=12
所以√3<b+c<=2√3
(b-2c)cosA=a-2acos^2(B/2)
(b-2c)cosA=a-a(1+cosB)=-acosB
由a/sinA=b/sinB=c/sinC=R,两边除R,则:
sinBcosA-2sinCcosA=-sinAcosB
sin(B+A)=2sinCcosA
由C=π-(B+A),则sinC=sin(B+A)
sinC=2sinCcosA...
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(b-2c)cosA=a-2acos^2(B/2)
(b-2c)cosA=a-a(1+cosB)=-acosB
由a/sinA=b/sinB=c/sinC=R,两边除R,则:
sinBcosA-2sinCcosA=-sinAcosB
sin(B+A)=2sinCcosA
由C=π-(B+A),则sinC=sin(B+A)
sinC=2sinCcosA
cosA=1/2
A=π/3
a/sinA=b/sinB=c/sinC=(b+c)/(sinB+sinC)=√3/(√3/2)=2
b+c=2(sinB+sinC)=2[sinB+sin(2π/3-B)]=4sin(π/3)cos(B-π/3)=2√3cos(B-π/3)
0当B=π/3时,b+c最大值2√3
当B->0或B->2π/3时,b+c最小值√3
b+c的取值范围为(√3,2√3]
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