Question

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

Answer 1

设AC=BC=m

则AB=根号2 m

CD=BD=m/2

BE=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^2

CE=根号(5/8) m

CE/sinA=AC/sinAEC

sinAEC=AC/CE *sinA = m/[根号(5/8) m] *根号2/2 = 2根号5 /5