求大神解答这个高中数学题,百度复制的也行,我只要看得懂,谢谢了
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(1) 过B作BH垂直于AC交AC与点H
所以 c*cosA=AH
a*cosC=CH
c*cosA+ a*cosC=AH+CH=AC=b
2b*cosA=b
cosA=1/2
a=60
(2)A^C-A^B=B^C
|A^C-A^B|=BC=a=1
a/sinA=b/sinB=c/sinC
b=sinB/sin(60) c=sinC/sin(60)
B+C=120
l=a+b+c=1+(1/sin(60))*(sinB+sinC)
sinB+sinC=2sin((B+C)/2)cos((B-C)/2)=2*sin(30)*cos((B-C)/2)=cos((B-C)/2)
B+C=120 suoyi 1>=cos((B-C)/2)>1/2
suoyi
1>=sinB+sinC>1/2
suoyi
l=a+b+c=1+(1/sin(60))*(sinB+sinC)>1+2/根号3*(1/2)=1+根号3/3
l=a+b+c=1+(1/sin(60))*(sinB+sinC)<=1+2/根号3*1=1+2根号3/3
所以 c*cosA=AH
a*cosC=CH
c*cosA+ a*cosC=AH+CH=AC=b
2b*cosA=b
cosA=1/2
a=60
(2)A^C-A^B=B^C
|A^C-A^B|=BC=a=1
a/sinA=b/sinB=c/sinC
b=sinB/sin(60) c=sinC/sin(60)
B+C=120
l=a+b+c=1+(1/sin(60))*(sinB+sinC)
sinB+sinC=2sin((B+C)/2)cos((B-C)/2)=2*sin(30)*cos((B-C)/2)=cos((B-C)/2)
B+C=120 suoyi 1>=cos((B-C)/2)>1/2
suoyi
1>=sinB+sinC>1/2
suoyi
l=a+b+c=1+(1/sin(60))*(sinB+sinC)>1+2/根号3*(1/2)=1+根号3/3
l=a+b+c=1+(1/sin(60))*(sinB+sinC)<=1+2/根号3*1=1+2根号3/3
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谢谢啦^ω^
我一起还发了几道题,你看你会不呀,回答了秒采纳~
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