Question

三角形ABC中角A,B,C(C为钝角)所对的边分别为a,b,c,且COS(A+B-C)=1/4.a=2,sin(A+B)

Answer 1

由正弦定理,c/a=sinC/sinA=sin(A+B)/sinA=2,c=4;

cos(A+B-C)=cos(π-2C)=-cos(2C)=-2(cosC)^2+1=1/4,cosC=-sqrt(6)/4;

cosC=(a^2+b^2-c^2)/2ab,b=2*sqrt(6);