Question

已知三角形abc的周长为根号3+1，且sina+sinb=根号3sinc,三角形abc的面积为8分之3sinc,

求tan《a+b》的值

Answer 1

∵在△ABC中S=（3/8)SinC 又 ∵S=(1/2)abSinC∴ab=3/4
∵SinA+SinB=√3SinC又 ∵a/SinA=b/SinB=c/SinC=2R∴a/2R+b/2R=√3c/2R∴a+b= √3c
∵a+b+c=√3+1∴√3c+c=√3+1∴c=1
∵a+b=√3∴a2+b2=3/2
∵在△ABC中tan(a+b)=tan(180-c）=-tanC 又 ∵tanc=√(1/(CosC)2-1)
CosC=(a2+b2-c2)/2ab=1/3 ∴tanC=2√2 ∴tan(a+b)=-2√2