AB是圆O的直径BC垂直于AB于若sin角BAD=4比5,圆半径为5求DFB.连接OC交圆O于点E弦AD平行于OC弦DF

推测一下,是弦DF垂直BC于G连接BD,AB为直径,则∠ADB=90度在直角三角形中,BD=AB*sin∠BAD=10×4/5=8勾股定理,AD=6设∠BOD=a∠CBF是弦切角所以∠CBF=1/2∠BOD=1/2a∠DFB=180-∠CBF=180-1/2acos∠DFB=cos(180-1/2a)=-cos(1/2a)过O作OK垂直BD于K则OK为三角形ABD的中位线,OK=1/2AD=3cos(1/2a)=OK/OB=3/5所以cos∠DFB=-3/5