sinA 根号2sinB=2sinC,COS

∵根号3cosA=根号2cosBAB为三角形内角∴角A、B均为锐角∵sinA=√2sinB①,cosA=√(2/3)cosB②∴①^2+②^2：1=2sin^2(B)+2/3cos^2(B)=4/3s

sinA+sinB=√2sinCa/sinA=b/sinB=c/sinC有：(a+b+c)/(sinA+sinB+sinC)=(a+b)/(sinA+sinB)=c/sinC所以有：(√2+1-c)/

(sina+sinb)²=2,(cosa+cosb)²=4/3,前后两式相加得：2+2(cosacosb+sinasinb)=10/3,cos(a-b)=2/3；后式减前式得：2(

sinB+sinc=√2sinA,而用a/sinA=b/sinb=c/sinc=2R.代入得到b+c=√2a,a+b+c=√2+1.得a=1三角形ABC面积为1/6*sinA.知道bc=1/3有知道b

等于1啦设A,B,C三个角对应的边为La,Lb,LcLa=(SINA*Lc)/SINC(1)Lb=(SINB*Lc)/SINC(2)所以La+Lb+Lc=(1)+(2)+Lc=根号2+1其中SINA+

令cosA+cosB=x---(1)sinA+sinB=√2/2---(2)由两式平方相加的2+2cos(A-B)=x^2+1/2得-1≤cos(A-B)=x^2/2-3/4≤1可得x^2≤7/2,从

sin(a-B)cosa-1/2[sin(2a+B)-sinB]=sin(a-B)cosa-1/2[2cos(a+b)sina]=sin(a-b)cosa-cos(a+b)sina=sinacosbc

由正弦定理,a:b:c=√5：√35：2√5=1：√7：2,∴cosB=(1+4-7)/4=-1/2,∴B=120°,为所求.

sinA比sinB=根号2比1,a比b=根号2:1a方=2b方cosA×2bc=b方+c方-a方=c方-b方c方=b方+根号2bccosA×2bc=根号2bccosA=sinA=根号2/2sinB=1

sina+sinb=根号2……①cosa+cosb=(根号2)/3……②①^2+②^2,得(sina)^2+(sinb)^2+2sinasinb+(cosa)^2+(cosb)^2+2cosacosb

tana=（√3)tanb.1）2sin^2b+(2/3)cos^2b=12-2cos^2b+(2/3)cos^2b=13/4=cos^2bcosb=±√3/2由1）知,a,b属于I、III象限b=π

为了表述清楚,将a、b置换成A、B由sinA+sinB=√2可得：(sinA+sinB)²=sin²A+sin²B+2sinAsinB=1-cos²A+1-co

sin^2a=2sin^2b,cos^2a=2/3cos^2b,两式相加1=2sin^2b+2/3cos^2b,1=2(1-cos2b)/2+2/3(1+cos2b)/2cos2b=1/22b=π/3

答：sina+sinb=√2/2两边平方得：sin²a+2sinasinb+sin²b=1/2…………（1）设cosa+cosb=m两边平方得：cos²a+2cosaco

sinA+sinB=根号2/2(sina+sinb)²=1/2sin²a+2sinasinb+sin²b=1/2令k=cosa+cosbcos²a+2cosac

/>(sina-sinb)²=2/3sin²a+sin²b-2sinasinb=2/3(1)(cosa+cosb)²=1/2cos²a+cos

sinA+sinB=2sin[(A+B)/2]cos[(A-B)/2],cosA+cosB=2cos[(A+B)/2]cos[(A-B)/2],两式相除即得：tan[(A+B)/2]=(sinA+si

sinA+sinB=√2sinC,根据正弦定理a/sinA=b/sinB=c/sinC,所以sinA=asinC/csinB=bsinC/c(a+b)sinC/c=√2sinC,即a+b=c√2又因为

∵a,B均为锐角∴0

由正弦定理a/sinA=b/sinB=c/sinCsinA+sinB=√2sinC所以a+b=√2ca+b+c=2√2+2所以√2c+c=2√2+2所以AB=c=2a+b=√2c=2√2S=1/2ab