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解:(1)設(shè)直線AB的解析式為y=kx+b,
將A(0,2
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),B(2,0)代入解析式y(tǒng)=kx+b中,得
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,
解得:
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.
∴直線AB的解析式為y=-
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x+2
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,
將D(-1,a)代入y=-
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x+2
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得:a=3
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,
∴點(diǎn)D坐標(biāo)為(-1,3
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),
將D(-1,3
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)代入y=
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中得:m=-3
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,
∴反比例函數(shù)的解析式為y=-
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;
(2)聯(lián)立得:
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,
解得:
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或
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,
∴點(diǎn)C坐標(biāo)為(3,-
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),
過(guò)點(diǎn)C作CM⊥x軸于點(diǎn)M,則在Rt△OMC中,CM=
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,OM=3,
∴tan∠COM=
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=
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,∴∠COM=30°,
在Rt△AOB中,tan∠ABO=
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=
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=
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,
∴∠ABO=60°,
∴∠ACO=∠ABO-∠COE=30°;
(3)如圖,若OC′⊥AB,則有∠BNO=90°,
∵∠NBO=60°,∴∠BON=30°,
∵∠COM=30°,
∴∠COC′=∠COM+∠BON=60°,即旋轉(zhuǎn)角為60°,
則當(dāng)α=60°時(shí),OC′⊥AB.
故答案為:60.
分析:(1)設(shè)直線AB解析式為y=kx+b,將A與B坐標(biāo)代入求出k與b的值,確定出直線AB解析式,將D坐標(biāo)代入直線AB解析式求出a的值,確定出D坐標(biāo),將D坐標(biāo)代入反比例解析式求出m的值,即可確定出反比例解析式;
(2)聯(lián)立直線AB與反比例解析式,求出交點(diǎn)C坐標(biāo),過(guò)C作CM垂直于x軸,在直角三角形COM值,利用銳角三角函數(shù)定義及特殊角的三角函數(shù)值求出∠COM的度數(shù),在直角三角形AOB中,同理求出∠ABO的度數(shù),由外角性質(zhì)即可求出∠ACO的度數(shù);
(3)根據(jù)題意畫(huà)出圖形,求出OC′⊥AB時(shí)的旋轉(zhuǎn)角即可確定出α度數(shù).
點(diǎn)評(píng):此題屬于反比例函數(shù)綜合題,涉及的知識(shí)有:待定系數(shù)法確定函數(shù)解析式,一次函數(shù)與反比例函數(shù)的交點(diǎn)問(wèn)題,銳角三角函數(shù)定義,以及旋轉(zhuǎn)的性質(zhì),熟練掌握待定系數(shù)法是解本題的關(guān)鍵.