:因为∠ACB=90°,所以∠A+∠B=90°.因为CD⊥AB,所以∠B+∠BCD=90°,所以∠BCD=∠A,所以tan∠BCD=tanA==.Р10.解:∵=,∴∠AED=∠ABC.Р在Rt△ABC中,AC=1,AB=2,Р由勾股定理得BC=.Р∴cos∠AED=cos∠ABC==.Р11.解:(1)证明:∵AB为⊙O的直径,Р∴∠ACB=90°.Р又∵OD∥BC,Р∴∠AEO=∠ACB=90°,Р∴OD⊥AC,Р∴=,Р∴AD=CD.Р(2)∵AB=10,Р∴OA=OD=AB=5.Р∵OD∥BC,Р∴∠AOE=∠ABC.Р在Rt△AEO中,OE=OA·cos∠AOE=OA·cos∠ABC=5×=3,Р∴DE=OD-OE=5-3=2.Р由勾股定理,得AE===4.Р在Rt△AED中,tan∠DAE===.Р又∵∠DBC=∠DAE,Р∴tan∠DBC=.Р12.解:如图,过点A作AD⊥l1于点D,交l2于点F,过点B作BE⊥l1于点E,设l1和l2之间的距离为1,则l2和l3之间的距离也为1.Р∵∠CAD+∠ACD=90°,∠BCE+∠ACD=90°,Р∴∠CAD=∠BCE.Р∵△ABC是等腰直角三角形,∴AC=BC.Р在△ACD和△CBE中,Р,Р∴△ACD≌△CBE(AAS),Р∴CD=BE=1.Р在Rt△ACD中,AC==,Р在等腰直角三角形ABC中,РAB==.Р∵l2∥l3,∴∠ABF=∠α,Р∴sinα=sin∠ABF===.Р13.30Р14.解:原式=1-×=-.Р15.解:sin21°+sin22°+sin23°+…+sin287°+sin288°+sin289°=(sin21°+sin289°)+(sin22°+sin288°)+(sin23°+sin287°)+…+(sin244°+sin246°)+sin245°=1+1+…+1+0.5=44.5.
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