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

2025-05-22 10:05:42
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回答1:

∵∠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