Question

如图,Rt△ABC中,∠C=90°,AD平分∠BAC,若AD=5,AC=4,求∠B的度数？

Answer 1

在Rt△ACD中,∠C=90°,AD=5,AC=4,
根据勾股定理得：CD=√(AD²-AC²)=3
∴CD/AD=3/5,
∴sin∠CAD=3/5 cos∠CAD=4/5,
又AD为∠BAC的平分线,
∴∠CAD=∠BAD,即∠CAB=2∠CAD,
则∠B=90°-∠CAB=90°-2∠CAD；
cos(∠B)=cos(90°-2∠CAD)=sin(2∠CAD)=2sin(∠DAC)cos(∠DAC)=2*(3/5)(4/5) =24/25；
∠B=arccos(24/25)≈16°,10,cos(∠DAC)=AC/AD=0.8
DC^2+AC^2=AD^2，所以DC=3
sin(∠DAC)=AC/AD=0.6
cos(∠B)=sin(∠BAC)=sin(2*∠DAC)=2sin(∠DAC)cos(∠DAC)=0.96
∠B=arccos(0.96)≈16°,2,图呢。。。。,0,