Question

△ABC中,a+b=7,并且4sin^2(A+B)/2+cos2C=1/2,求∠C大小 求△ABC的面积

Answer 1

sin(A+B)/2=sin(180°-C)/2=sin(90°-C/2)=cos(C/2) ∴4sin(A+B)/2+cos2C=4cos(C/2)+cos2C=4×(1+cosC)/2+2cosC-1=1/2 整理得 2cosC+2cosC+1/2=2(cosC+1/2)=0,cosC=-1/2,∴∠C=120° 余弦定理,cosC=(a+b-c)/2ab=[(a+b)-c-2ab]/2ab=(49-37-2ab)/2ab=(6-ab)/ab=-1/2 ∴ab=12,S△ABC=(1/2)absinC=(1/2)×12×(√3/2)=3√3