已知△ABC的三个内角A,B,C所对应的边分别为a,b,c,且tanBtanC-√3(tanB+tanC)=1.(1)求∠A (2)现给出三个条件:①a=1②b=2sinB③2c-(√3+1)b=0,请从中选出两个条件切△ABC的面积是求△ABC的面积
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![已知△ABC的三个内角A,B,C所对应的边分别为a,b,c,且tanBtanC-√3(tanB+tanC)=1.(1)求∠A (2)现给出三个条件:①a=1②b=2sinB③2c-(√3+1)b=0,请从中选出两个条件切△ABC的面积是求△ABC的面积](/uploads/image/z/2465177-41-7.jpg?t=%E5%B7%B2%E7%9F%A5%E2%96%B3ABC%E7%9A%84%E4%B8%89%E4%B8%AA%E5%86%85%E8%A7%92A%2CB%2CC%E6%89%80%E5%AF%B9%E5%BA%94%E7%9A%84%E8%BE%B9%E5%88%86%E5%88%AB%E4%B8%BAa%2Cb%2Cc%2C%E4%B8%94tanBtanC-%E2%88%9A3%28tanB%2BtanC%29%3D1.%281%29%E6%B1%82%E2%88%A0A+%282%29%E7%8E%B0%E7%BB%99%E5%87%BA%E4%B8%89%E4%B8%AA%E6%9D%A1%E4%BB%B6%EF%BC%9A%E2%91%A0a%3D1%E2%91%A1b%3D2sinB%E2%91%A22c-%28%E2%88%9A3%2B1%29b%3D0%2C%E8%AF%B7%E4%BB%8E%E4%B8%AD%E9%80%89%E5%87%BA%E4%B8%A4%E4%B8%AA%E6%9D%A1%E4%BB%B6%E5%88%87%E2%96%B3ABC%E7%9A%84%E9%9D%A2%E7%A7%AF%E6%98%AF%E6%B1%82%E2%96%B3ABC%E7%9A%84%E9%9D%A2%E7%A7%AF)
已知△ABC的三个内角A,B,C所对应的边分别为a,b,c,且tanBtanC-√3(tanB+tanC)=1.(1)求∠A (2)现给出三个条件:①a=1②b=2sinB③2c-(√3+1)b=0,请从中选出两个条件切△ABC的面积是求△ABC的面积
已知△ABC的三个内角A,B,C所对应的边分别为a,b,c,且tanBtanC-√3(tanB+tanC)=1.(1)求∠A (2)现给出三个条件:①a=1②b=2sinB③2c-(√3+1)b=0,请从中选出两个条件切△ABC的面积
是求△ABC的面积
已知△ABC的三个内角A,B,C所对应的边分别为a,b,c,且tanBtanC-√3(tanB+tanC)=1.(1)求∠A (2)现给出三个条件:①a=1②b=2sinB③2c-(√3+1)b=0,请从中选出两个条件切△ABC的面积是求△ABC的面积
1.
tanBtanC-√3(tanB+tanC)=1
1-tanBtanC=-√3(tanB+tanC)
(tanB+tanC)/(1-tanBtanC)=-√3/3
tan(B+C)=-√3/3
B+C=150°
∠A=30°
2.原来这题是接着上题的,我说怎么少个条件.
选②③
∵∠A=30°
∴sinA=1/2,cosA=√3/2
∵a/sinA=b/sinB=2
∴a=1
∵2c-(√3+1)b=0
∴c=(√3+1)b/2 ,c²=[(√3+1)b/2]²=(√3/2 + 1)b²
a²=b²+c²-2bccosA=1
b²+(√3/2 + 1)b²-√3b[(√3+1)b/2]=1
b²=2,b=√2
c=(√6+√2)/2
S=bcsinA *(1/2)=(√3+1)/4