AB是圆O的直径,弦AC,BD相交于点P,若AB=3,CD=1,cos角DAP=?

问题描述:

AB是圆O的直径,弦AC,BD相交于点P,若AB=3,CD=1,cos角DAP=?

∵∠PAB=∠PDC∠PBA=∠PCD
∴△ABP∽△DCP
∴AB/CD=AP/DP=3/1=3
∵AB是圆O的直径
∴∠ADP=90°
sin∠DAP=DP/AP=1/3
cos∠DAP=√[1-(sin∠DAP)^2]=2√2/3