在三角形ABC中,三个角都为锐角, 若角A=60度,a=根号3,求b+c的 范围在线等,很急.
来源:学生作业学帮网 编辑:学帮网 时间:2024/05/23 23:25:22
在三角形ABC中,三个角都为锐角, 若角A=60度,a=根号3,求b+c的 范围
在线等,很急.
根据正弦定理,a/sinA=b/sinB=c/sinC,
a/sinA=(b+c)/(sinB+sinC)
b+c=[√3/(√3/2)]*2sin[(B+C)/2]*cos[(B-C)/2]
=4sin[(180°-A)/2]cos[(B-C)/2]
=4sin(90°-A/2)cos[(120°-C-C)/2]
=4cos(A/2)cos(60°-C)
=4*cos30°*cos(60°-C),
=2√3cos(60°-C),
0
根据正弦定理,a/sinA=b/sinB=c/sinC,
a/sinA=(b+c)/(sinB+sinC)
b+c=[2√3/(√3/2)]*2sin[(B+C)/2]*cos[(B-C)/2]
=8sin[(180°-A)/2]cos[(B-C)/2]
=8sin(90°-A/2)cos[(120°-C-C)/2]
=8cos(A/2)cos(60°-C)...
全部展开
根据正弦定理,a/sinA=b/sinB=c/sinC,
a/sinA=(b+c)/(sinB+sinC)
b+c=[2√3/(√3/2)]*2sin[(B+C)/2]*cos[(B-C)/2]
=8sin[(180°-A)/2]cos[(B-C)/2]
=8sin(90°-A/2)cos[(120°-C-C)/2]
=8cos(A/2)cos(60°-C)
=8*cos30°*cos(60°-C),
=4√3cos(60°-C),
0
收起
http://zhidao.baidu.com/question/261096795.html
对任意三角形, 都有下式成立:
a / sinA = b / sinB = c / sinC
此题中, a = 根号3, sinA = 根号3 / 2
故
b / sinB = c / sinC = 2.
从而
b = 2sinB, c = 2sinC
b+c = 2(sinB + sinC) = 2 * 2sin( (B+C)/2...
全部展开
对任意三角形, 都有下式成立:
a / sinA = b / sinB = c / sinC
此题中, a = 根号3, sinA = 根号3 / 2
故
b / sinB = c / sinC = 2.
从而
b = 2sinB, c = 2sinC
b+c = 2(sinB + sinC) = 2 * 2sin( (B+C)/2 ) * cos( (B-C)/2 )
注意 B + C = 120度
故
b + c = 2*根号3 * cos( (B-C)/2 )
而 0 <= B-C < 90度, 从而 根号2 / 2 < cos( (B-C)/2 ) < 1
最后
根号6 < b + c < 2*根号3
收起
﹙√3,3﹚