Question

在△ABC中,sinB=2√2/3,b=4√2,a=c,求△ABC的面积

Answer 1

cosB=±√［1-(sinB)^2］=±1/3,

b^2=a^2+c^2-2accosB,

∴32=2a^2±2/3a^2,

4a^2=96或8a^2=96,

a^2=24或a=12,

∴S=1/2a^2sinB=16√2或

S=1/2×12×2√2/3=4√2.