Question

在△ABC中,c＝2又根号2,a＞b,C＝π／4,且有tanAtanB＝6,试求a,b及此时三角形的面积.

Answer 1

tanB=6/tanA tanB=tan(π-A-C)=-tan(A+C)=-(tanA+tanC)/(1-tanAtanC)=(tanA+1)/(tanA-1) ∴tan^2A+tanA=6tanA-6 tan^2A-5tanA+6=0 tanA=2或3 对应的tanB=3或2 则A,Bb sinA>sinB A>B tanA>tanB tanA=3,tanB=2 ∴sinA=3/√10,sinB=2/√5 正弦定理：a:b:c=sinA:sinB:sinC=3:2√2:√5 ∴a=3/√5*2√2=6/5*√10 b=2√2/√5*2√2=8/5*√5 S=1/2*6/5*√10*8/5*√5*√2/2=24/5