三角形ABC中,角A,B,C对应的边分别是a,b,c,若sinA,sinB,sinC成等差数列,且tanC=2√2.(1)求sinB/sinA

sinA,sinB,sinC成等差数列,且tanC=2根号2,则有sinC=2/3根号2,cosC=1/3即有2sinB=sinA+sinC,即有2b=a+ccosC=(a^2+b^2-c^2)/(2ab)=(a^2+b^2-(2b-a)^2)/(2ab)=(a^2+b^2-4b^2+4ab-a^2)/(2ab)=(4ab-3b^2)/(2ab)=2-3b/(2a)...若c=11,求三角形ABC面积b=10/9a2b=a+11 20/9a=a+11a=9,b=10S＝1/2absinC=1/2*9*10*2/3根号2＝30根号2