4.19-3/在⊿ABC中,∟A,∟B,∟C对边的长分别是a,b,c,
来源:百度知道 编辑:UC知道 时间:2024/05/20 18:07:42
求:m的值。
题目改一下^_^
4.19-3/在⊿ABC中,∟A,∟B,∟C对边的长分别是a,b,c,且有:a^2+b^2=m*c^2(m为常数),若(cotC)/(cotA+cotB)=999+(1/2)
求:m的值。
请写出详细解题过程及作同类题目的思路,谢~~
cotC/(cotA+cotB) = (cosC/sinC)/[cosA/sinA + cosB/sinB] =
(cosC/sinC)/[sin(A+B)/sinAsinB] =
cosC * sinA * sinB / (sinC * sinC), 这里用到了 A+B+C = 180
利用 a/sinA = b/sinB = c/sinC
上式 = cosC * ab/c^2 = 999.5
另一方面,
a^2 + b^2 - 2ab*cosC = c^2
a^2 + b^2 - c^2 = 2ab * cosC
(m-1)c^2 = 2ab * cosC
m-1 = 2ab * cosC / c^2 = 2*999.5 = 1999
m = 2000
cotC/(cotA+cotB)=999+(1/2)...............(1)
a^2+b^2=mc^2..............................(2)
cotA+cotB=cosA/sinA+cosB/sinB=(cosAsinB+sinAcosB)/sinAsinB
=sin(A+B)/sinAsinB=sinC/sinAsinB
cotC=cosC/sinC
所以cotC/(cotA+cotB)=cosCsinAsinB/(sinC)^2.....(3)
因为cotC/(cotA+cotB)=999+(1/2)
所以cosCsinAsinB/(sinC)^2=999+(1/2).........(4)
由余弦定理c^2=a^2+b^2-2abcosC a^2+b^2=c^2-2abcosC代入(2)
c^2-2abcosC=mc^2
即(m-1)c^2=2abcosC.......................(5)
由正弦定理a/sinA=b/sinB=c/sinC=2R (R为△外接圆半径)得
a=2RsinA b=2RsinB c=2RsinC代入(5)整理
(m-1)(sinC)^2=