Question

设△ABC中三内角A,B,C所对的边分别为a,b,c,已知cosB=3/5且向量AB*BC=-21

求△ABC的面积S

Answer 1

解
AB×BC
=/AB//BC/cos(π-B)
=-/AB//BC/cosB
=-/AB//BC/×3/5
=-21
∴/AB//BC/=35
又cosB=3/5
B∈(0,π)
∴sinB>0
∴sinB=4/5
∴S=1/2/AB//BC/sinB
=1/2×35×4/5
=14

Answer 2

cosB是-3/5吧，如果是就是
|AB|*|BC|*cosB=-21得|AB|*|BC|=35
cosB=-3/5
sinB=4/5得S=1/2*|AB|*|BC|*sinB=14