Question

在△ABC中三边a,b,c所对的角分别为A,B,C,已知a＝3√2,b＝√10,c＝2,求角B的度数和△ABC的面积

Answer 1

作AD⊥BC,垂足D,设BD=x,

AB^2-BD^2=AD^2=AC^2-DC^2

4-x^2=10-(3√2-x)^2

x=√2,

AD^2=AB^2-BD^2=4-2=2

AD=√2=BD,角B=45°；

S△ABC=BC*AD/2=3√2*√2/2=3.