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