Question

三角形ABC中,角A,B,C对应的边分别是abc,若sinA,sinB,sinC成等差数列,且tanC=2根号2. 求s

Answer 1

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+c

cosC=(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)=1/3

即有3b/(2a)=5/3

b/a=10/9

即有sinB/sinA=b/a=10/9