试题分析:(1)设直线AB的解析式为
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,
将A(0,2
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),B(2,0)代入解析式
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中,得
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,解得
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。
∴直线AB的解析式为
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。
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将D(-1,
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)代入
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得,
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。
∴点D坐标为(-1,
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)。
将D(-1,
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)代入
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中得,
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。
∴反比例函数的解析式为
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。
(2)解方程组
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得
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,
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。
∴点C坐标为(3,
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),
过点C作CM⊥
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轴于点M,则在Rt△OMC中,
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,
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,∴
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,∴
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。
在Rt△AOB中,
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=
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,∴
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。
∴∠ACO=
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。
(3)如图,∵OC′⊥AB,∠ACO=30°,
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∴
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= ∠COC′=90°-30°=60°,∠BOB′=
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=60°。
∴∠AOB′=90°-∠BOB′=30°。
∵∠OAB=90°-∠ABO=30°,∴∠AOB′=∠OAB,
∴AB′= OB′=2.
答:当α为60度时OC′⊥AB,此时线段AB′的长为2。
点评:本题考查求函数解析式、三角函数,要求考生会用待定系数法求函数的解析式,掌握三角函数的定义,会用三角函数解题