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，目前答案易对网已经整理了衡水名师卷·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，更多联考试卷请关注本网站。

(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

One possible version:Dear Jack,The Dragon Boat Festival,which falls on the 5th day of the 5th lunar month,is one of the most im-portant traditional festivals.Every year Chinese people will hold grand celebrations in memory of QuYuan.Knowing that you are interested in Chinese traditions,I'm writing to invite you to take part in thecelebration to be held by our school.The celebration will take place by the lake of our school on Friday.In the morning,we will wrapand cook "zongzi",which will be eaten as lunch.In the afternoon,a dragon boat race will be held on thelake.I'm sure you will have a good time.Waiting for your coming.Yours,Li Hua