如图三角形ABC是等腰三角形∠ACB=90°，过BC中点D作DE垂直AB垂足E连接CE求sin∠AEC

显示全部楼层 · 2011-2-20 12:13:27

解：过C作AB的垂线，交AB于F。设DE=X。由题意知，△BCF和△BDE均为等腰直角三角形，BF=CF，BE=DE，DE为△BCF的中位线，∴CF=2DE=2X，EF=BE=DE=X∴CE=（CF^2+EF^2)^(1/2)=((2X)^2+X^2)^(1/2)=5^(1/2)X∴sin∠AEC=CF/CE=2X/(5^(1/2)X)=2√5/5

千问 · 2011-2-20 12:13:27

设AC=BC=m则AB=根号2 mCD=BD=m/2BE=BDcos45°=m/(2根号2)CE^2=BC^2+BE^2-2BC*CEcos45°=m^2+m^2/8-2m*m/(2根号2)*根号2/2=5/8 m^2CE=根号(5/8)mCE/sinA=AC/sinAECsinAEC=AC/CE *sinA = m/[根号(