AB是圆o的直径,AB=10,CD是圆o的弦,AD与BC相交于P,若CD=6,COS角APC?

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AB是圆o的直径,AB=10,CD是圆o的弦,AD与BC相交于P,若CD=6,COS角APC?
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AB是圆o的直径,AB=10,CD是圆o的弦,AD与BC相交于P,若CD=6,COS角APC?
AB是圆o的直径,AB=10,CD是圆o的弦,AD与BC相交于P,若CD=6,COS角APC?


AB是圆o的直径,AB=10,CD是圆o的弦,AD与BC相交于P,若CD=6,COS角APC?
连接OC,OD,作OH垂直于CD,交CD于H,H为CD中点,CH = HD = 3,OC = OD = 5
∠APC = ∠PCD + ∠CDP
= ∠BCD + ∠CDA
= ∠BOD / 2 + ∠COA / 2 (圆周角等于圆心角的一半)
= ( π - ∠COD ) / 2
= π / 2 - ∠COH (∠COH = ∠DOH = ∠COD / 2)
所以
cos ∠APC = cos (π / 2 - ∠COH) = sin ∠COH = CH / CO = 3 / 5