Question

如图，已知在△ABC中，点D是BC上一点,AC=BD, ∠CAD=30°，∠ACB=40°，求∠ABC的度数.

Answer 1

运用三角法求解，根据正弦定理
△ABD中，BD/sin∠BAD=AD/sin∠B……①
△ADC中，AC/sin∠ADC=AD/sin∠C……②
因为∠C=40°，∠ADC=180°-∠CAD-∠C=180°-30°-40°=110°
∠BAD=∠ADC-∠B=110°-∠B
且AC=BD
代入①②两式，得：
AC/sin(110°-∠B)=AD/sin∠B
AC/sin110°=AD/sin40°
上述两式相除，得：
sin110°/sin(110°-∠B)=sin40°/sin∠B
sin70°/sin(70°+∠B)=sin40°/sin∠B
sin70°/sin40°=(sin70°cos∠B+cos70°sin∠B)/sin∠B=sin70°cot∠B+cos70°
cot∠B=1/sin40°-cot70°
=(sin70°-cos70°sin40°)/sin40°sin70°
=(sin110°-cos70°sin40°)/sin40°sin70°
=[sin(70°+40°)-cos70°sin40°)/sin40°sin70°
=sin70°cos40°/sin40°sin70°
=cot40°
所以∠B=40°