Question

在△ABC中,已知AB=根号3,BC=2,cosB=-根号3/6,求sinC的值

Answer 1

cosB=(a²+c²-b²)/(2ac)=(3+4-b²)/(4√3)=-√3/6;

7-b²=-2;

b²=9;

b=3;

cosC=(a²+b²-c²)/(2ab)=(4+9-3)/(2*2*3)=10/12=5/6;

sinC=√(1-25/36)=√11/6;

0＜C＜π/2;