衡水名师卷·2022-2023学年度高三分科检测提分卷·151靶向提升(老高考)文综(一)1试题 答案

衡水名师卷·2022-2023学年度高三分科检测提分卷·151靶向提升(老高考)文综(一)1试题 答案，目前衡水金卷先享题网已经汇总了衡水名师卷·2022-2023学年度高三分科检测提分卷·151靶向提升(老高考)文综(一)1试题 答案的各科答案和试卷，更多衡水金卷答案请关注本网站。

I go for a walk in the park near my home on every day.Yesterday,while taking a walk,I see asawboy having a picnic with his parents.We ate snacks and cakes with rubbish left on a ground.WhenTheythethey A about to leave,the boy parents noticed the rubbish.They told the boy how rubbish harmed thewereboy'senvironment.Heard what his parents said,the boy instant picked up all the rubbish.We should learnHearinginstantlyfrom the family or try our best to protect the environment.Only in this way can we keep the earth cleanandandbeautybeautiful

(1)证明：因为直线AB⊥平面BCDE,所以AB⊥BC,AB⊥BE,因为底面BCDE是梯形，BC∥DE,BC⊥CD,CD=DE=2BC=2,F是边BC的中点，所以四边形CDEF为正方形，BE=√EF2十BF=2√2，CE2=8,AE2=AB2+BE2=AB2+8,AC2=AB2+BC2=AB2+16,所以AC2=AE2十CE2,所以AE⊥EC.(2)解：以FE,FC所在直线分别为x,y轴，在平面ABC中，过点F且垂直于FC的直线为之轴，建立如图所示的空间直角坐标系Fxyz,设AB=a,则C(0,2,0),E(2,0,0),D(2,2,0),A(0,-2,a),FA=(0,-2,a),FD=(2,2,0),设平面ADF的一个法向量为n=(x,y,z),FA·n=-2y十az=0,则所以n=(-a,a,2).FD·n=2x+2y=0,E因为直线AB⊥平面BCDE,所以AB⊥CE,又EC2十EB2=BC2,所以BE⊥EC,因为BE∩AB=B,所以CE⊥平面ABE,所以平面ABE的一个法向量为CE=(2,一2,0)，因为平面ADF与平面ABE所成二面角为45°，所以cos(CE,n)1=cE·nAa=c0s45°=CE1|n|22X√2a2+4,解得a=√2，所以AD=(2,4,一√2).设直线AD与平面ABE所成的角为0，则sin0=cos(AD,C克1=1A方.CE4ADICE22X2211所以直线AD与平面ABE所成角的正弦值为√1I11