Question

如图,在Rt△ABC中,∠B=90°,AB=BE=EF=FC.求∠AFE+∠ACE的度数.

Answer 1

设AB=m，则AF²=AB²+BF²=m²+(2m)²=5m², AF=√5m,
AC²=AB+²BC²=m²+(3m)²=10m², AC=√10m
∴sin∠AFE=AB/AF=m/√5m=√5/5,
cos∠AFE=BF/AF=2m/√5m=2√5/5
sin∠ACE=AB/AC=m/√10m=√10/10
cos∠ACE=BC/AC=3m/√10m=3√10/10
∵sin(∠AFE+∠ACE)=sin∠AFE*cos∠ACE+cos∠AFE*sin∠ACE=(√5/5)*(3√10/10)+(2√5/5)*(√10/10)=√2/2
∴∠AFE+∠ACE=45°

Answer 2

解：设∠ACE=α，∠AFE=β
则：tgα=1/3，tg β=1/2
根据两角和的正切公式tg(α+β)=(tgα+tgβ)/(1-tgα tgβ)求得：tg(α+β)=25/36≈0.6944
差正切表可求得

Answer 3

∠AFE+∠ACE=45°