第1个回答 2012-04-08
解:由ABC成等差数列,则有2B=A+C,又A+B+C=π,所以,B=π/3
①由正弦定理得:a/sinA=b/sinB=c/sinC,所以,sinA=asinB/b=3*(√3/2)/√13=3√39/26
由于,a=3<b=√13,所以A<B=π/3,则,cosA=√(1-sin²A)=5√13/26
sinC=sin(π-A-B)=sin(A+B)=sinAcosB+cosAsinB
=(3√39/26)*(1/2)+5√13/26*(√3/2)
=2√39/13
所以,c=bsinC/sinB=√13*(2√39/13)/(√3/2)=4
②t=sinAsinC =sin(π-B-C)sinC=sin(2π/3-C)sinC
=(1/2)[cos(2C-2π/3)-cos2π/3](积化和差)
=(1/2)cos(2C-2π/3)+1/4
由于,0<C<2π/3,所以,π/6<2C+π/6<3π/2
所以当cos(2C-2π/3)=cos0=1时,C=π/3,t取得最大值
即t=3/4