化学CY5衡水金卷信息卷2025试卷答案

化学CY5衡水金卷信息卷2025试卷答案，30 天真题突击计划：拆解题型、逐个击破，零基础也能逆袭提 50 分。真题不是 “做完就扔”，而是 “反复研究”—— 错题重做 3 遍，同类题再也不错。

(2)如图，以点A为坐标原点，AB,AD所在直线分别为x,y轴建立如下图所示的空间直角坐标系，ZAPyDEC;GAFB交过点A在平面ABCD内作AG⊥EF,连接G,由已知可得AF=PF,AE=PEn AFG主PFG,FG=FG,口故△AFG≌APFd则∠PGANGF=30°,即PG⊥EF,且PG=AG,因为平面PEF⊥平面BCDEF,平面PEF∩平面BCDEF=EF,PGC平面PEF,所以PG⊥平面BCDEF,设EF=m,∠AEF=a(0°<a<90°),则AE=mcosa,AF=msin .因为AE=BF,  几则AB=Nm固E=msin a+mcos a=2,AG=AEsimjnsin acos a,易得G(msin acos²a,msin²acos a,0),则点P(msin acos²a,msin²acosa,msin acos a),E(0,mcosa,0),C(2,2,0),EC=(2,msin a,0),PE=(-msin acos²a,mcos³a,-msin acos a),设平面PEC的一个法向量为n=(x,y则-严台(12分)令x=—sim得  sn²a,sin²a+sin acos a,sin acos²aHcesa),易知平面ECF的一个法向量为u=(0,0,1),

cos(n,μu>=n”·“Vomae+(sme+san acs+Lo  s因为cos 75°=cos(45°+30°=s5°ps 30°—45 sn ao=5C令，    口可得sm“a+sm²a(sm2       cosa二2—√3-06(14分)所以(2+√3)²=sin²a(3¹as¹²sin asios acos a)_tan²a(2sim²st2sa sc sosco²2Ssa²a+co&2e)tan²ac ztat’ar-?tna+1(fana+1)易知函数g(x)在(0,十∞)上连续，因为g(1)=9—(7+4√3)<0,g(4)>0,由零点存在性定理可知，存在x。∈(1,4),使得g(xo)=0,因此，存在点E,使得二面角的P—CEF的平面角为75°.引远(17分)

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A lost Russian Blue cat named Luke has been reunitedwith his owners,Cindy and Jeff Hall,five years after hedisappeared from their Tucson home.

Twelve years ago,Cindy and Jeff Hall rescued Lukeas a kitten from a cat shelter.For seven years,Luke was acherished member of their family until one day when heslipped out and didn't return.Despite their best efforts,which included searching their neighborhood and postingflyers,Luke remained missing for years.Eventijallyhefamily moved to Nevada,never giving up hope bpt hbelievnethey might never see their beloved qat again.

However,last month,th ca sheller received an un-expected phone oall [Infondinyg them that Luke had beenfound at a gas saiod in Arkansas over 1,200 miles awayfrom his Tucson home.The shelter immediately contactedCindy and Jeff with the incredible news that their long-lostcat was not only alive but had been found.

化学CY5衡水金卷信息卷2025试卷答案，真题不是 “做完就扔”，而是 “反复研究”—— 错题重做 3 遍，同类题再也不错。30 天真题突击计划：拆解题型、逐个击破，零基础也能逆袭提 50 分。