【答案】
分析:(1)由于∠ABQ<90°,若△ABQ是直角三角形,需要考慮兩種情況:
①∠BAQ=90°,此時△BAQ∽△ABO,根據相似三角形所得比例線段,可求出BQ的長,即可得到Q點坐標;
②∠BQA=90°,此時四邊形BOAQ是矩形,BQ=OA,由此可求出Q點坐標.
(2)假設P點翻折到AB上時,落點為E,那么∠QAP=∠QAE,QE=QP;由于BQ∥OP,那么∠QAP=∠BQA=∠BAQ,即BQ=BA=5,此時P、Q運動了2.5s,所以AP=AE=
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,即E是AB的中點;分別過E、Q作BQ、OP的垂線,設垂足為F、H,易求EF=PH=
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,即可證得△QPH≌△QEF,得∠EQF=∠PQH,由此發(fā)現(xiàn)∠EQP=90°,而∠PQA=∠EQA,由此可求得∠AQP的度數(shù).
(3)假設存在這樣的平行四邊形,可分作兩種情況考慮:
①點C在線段PQ上,可延長AC、BQ交于點F,由于DQ∥AC,因此DQ是△BCF的中位線,則FC=2DQ=2AC,過F作FH⊥x軸于H,由于∠BAC=90°,可證得△AOB∽△FHA,通過得到的比例線段,即可求出AF的長,進而可得到AC的長;在Rt△BAC中,已知了AC、BA的長,即可求出∠ABC的正切值;
②點C在PQ的延長線上,設AD、AC與BQ的交點分別為G、F,按照①的思路可證得AD=CQ=2AG,那么在相似三角形△CFQ和△AFG中,F(xiàn)C=2AF,即AC=3AF,AF的長在①中已求得,由此可得到AC的長,進而可求出∠ABC的正切值.
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解:(1)根據題意,可得:A(4,0)、B(0,3),AB=5.
�。┊敗螧AQ=90°時,△AOB∽△BAQ,
∴
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.解得
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;
ⅱ)當∠BQA=90°時,BQ=OA=4,
∴Q
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或Q(4,3).(4分)
(2)令點P翻折后落在線段AB上的點E處,
則∠EAQ=∠PAQ,∠EQA=∠PQA,AE=AP,QE=QP;
又BQ∥OP,
∴∠PAQ=∠BQA,∴∠EAQ=∠BQA,
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即AB=QB=5.
∴
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,
∴
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,即點E是AB的中點.
過點E作EF⊥BQ,垂足為點F,過點Q作QH⊥OP,垂足為點H,
則
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,
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,∴EF=PH.
又EQ=PQ,∠EFQ=∠PHQ=90°,
∴△EQF≌△PQH
∴∠EQF=∠PQH,從而∠PQE=90°.
∴∠AQP=∠AQE=45°.(8分)
(3)當點C在線段PQ上時,延長BQ與AC的延長線交于點F,
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∵AC⊥AB,
∴△AOB∽△FHA.
∴
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即
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,
∴
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.
∵DQ∥AC,DQ=AC,且D為BC中點,
∴FC=2DQ=2AC.
∴
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.
在Rt△BAC中,tan∠ABC=
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;
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當點C在PQ的延長線上時,記BQ與AC的交點為F,記AD與BQ的交點為G,
∵CQ∥AD,CQ=AD且D為BC中點,
∴AD=CQ=2DG.
∴CQ=2AG=2PQ.
即:CQ:QP=2:1
又∵BQ∥OP
∴CF:AF=CQ:QP=2:1
∴FC=2AF,
又∵FA=
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,
∴FC=
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,
∴
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.
在Rt△BAC中,tan∠ABC=
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.(12分)
點評:此題考查的知識點較多,涉及到圖形的翻折變換、相似三角形及全等三角形的判定和性質、三角形中位線定理以及銳角三角函數(shù)的定義等知識,同時還考查了分類討論的數(shù)學思想,難度較大.