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