如图,AC⊥AB,BD⊥CD,AC与BD相交于点E,S△AED=25,S△BEC=36.求:cos∠AEB

来源:学生作业学帮网 编辑:学帮网 时间:2024/05/19 03:38:20

如图,AC⊥AB,BD⊥CD,AC与BD相交于点E,S△AED=25,S△BEC=36.求:cos∠AEB

因为 S△AED=1/2*EA*ED*sinAED=25,
S△BEC=1/2*EB*EC*sinBEC=36,两式相除得到:
(EA*ED*sinAED)/(EB*EC*sinBEC)=25/36.
因为 角AED=角BEC,EA/EB=cosAEB=cosDEC=ED/EC,所以有
(EA*ED*sinAED)/(EB*EC*sinBEC)
=(cosAEB)^2
=25/36
因为 角AEB是锐角,所以 cosAEB=5/6.