解:(1)∵OA=4
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,AB=4,∠OAB=90°,
∴tan∠AOB=
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,
∴∠AOB=30°,
∵OA=OH,OB=OB,∠BAO=∠BHO=90°,
∴Rt△AOB≌Rt△HOB(HL),
∴∠BOH=∠AOB=30°,
∴∠HOC=30°;
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(2)①過點N與H作NK⊥x軸,
∴NK∥OA,
∴△POQ∽△PKN,
∴當(dāng)
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=
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時,
∵OQ=4
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-t,OP=t,
∴PK=
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t,NK=
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(4
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-t),
∴OK=
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t,
∵∠HOC=30°,
∴
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,
∴t=
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,
∴當(dāng)t為
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時,QP=2PN;
②
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當(dāng)QP⊥OH時,△OPQ∽△HOB.
∵∠QPO=∠OHB=90°,∠QOP=∠OBH=60°,
∴△OPQ∽△HOB,
∴cos∠QOP=
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,
∴t=
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,
∴當(dāng)t=
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時,△OPQ與△HOB相似.
③當(dāng)PQ⊥OA時,△OPQ∽△BOH,
cos∠QOP=
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=
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,
解得:t=
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.
分析:(1)首先由三角函數(shù),求得∠AOB的度數(shù),由HL,可證得Rt△AOB≌Rt△HOB,即可求得∠HOC的度數(shù);
(2)首先作輔助線:過點N與H作NK⊥x軸,即可得到相似三角形:△POQ∽△PKN,由相似三角形的對應(yīng)邊成比例,即可求得t的值;
(3)由相似三角形的判定,易得當(dāng)QP⊥OH時,△OPQ∽△HOB,由三角函數(shù)的性質(zhì),即可求得當(dāng)t=
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時,△OPQ與△HOB相似.
點評:此題考查了相似三角形的判定與性質(zhì),以及三角函數(shù)的性質(zhì)與全等三角形的判定與性質(zhì).題目綜合性很強(qiáng),難度比較大,解題時要注意仔細(xì)分析求解.