英语周报高一北师大第32期答案

第二节书面表达【写作思路】根据试题要求,考生需要写一篇记叙文,记述身边一位值得尊敬和爱戴的人。首先考生需要介绍自己最尊敬和爱戴的人是谁,简单地介绍一下他(她)。然后,陈述为什么他(她)值得尊敬和爱戴。最后,可以对前文进行一个总结,使文章结构严谨。本文的写作重点在"尊敬和爱戴的原因",考生可围绕这个要点适当拓展,从性格特点、处事风格等方面展开写作。【范文赏读】The most beloved and respected person around me is my teacher, Ms LiThough she has been teaching English for twenty years, she is still passionateabout teaching. She is kind and considerate towards us just like our dear motherWe all respect her because she always tries new ways to make her classes livelyand interesting. Hardworking and knowledgeable, she is one of the best teachersin our school. When we have a problem, we will turn to her for help. She alwaystalks to us patiently and helps us to find a solution. In our eyes, she is not onlyour teacher, but also our best friend. Those are why she deserves our respect【亮点词句】亮点词汇: beloved, respected, be passionate about, be consideratetowards, turn to sb. for help, in one's eyes亮点句式:Why引导的表语从句( why she deserves our respect)

19.【考查目标】必备知识:本题主要考查直三棱柱中的线线垂直、二面角的正弦值、空间向量等知识.关键能力:通过线线垂直的证明和二面角的求解考查了空间想象能力、逻辑思维能力和运算求解能力学科素养:理性思维、数学探索【解题思路】(1)先证明BA⊥BC,再利用AB,BC,BB1两两垂直建立空间直角坐标系,求出相关点的坐标,利用向量证明;(2)分别求出面BB1C1C和面DEF的一个法向量,通过求出两法向量夹角的余弦值的最大值来解决解:(1)因为E,F分别是AC和C1的中点,且AB=BC=2所以CF=1,BF=√5.如图,连接AF,由BF⊥A1B1,AB∥AB,得BF⊥AB,于是AF=BF+ABF=3,所以AC=√AF-CF=2.由AB2+BC2=AC2,得BA⊥BC,故以B为坐标原点,以AB,BC,B1所在直线分别为x,y,z轴建立空间直角坐标系B-xy则B(0,0,0),E(1,1,0),F(0,2,1),BF=(0,2,1)设B1D=m(0≤m≤2),则D(m,0,2),于是DE=(1-m,1,-2)所以B.DE=0,所以BF⊥DE(2)易知面BB1C1C的一个法向量为n1=(1,0,0)设面DFE的法向量为n2=(x,y,2),n,又靂=(1-m,1,-2),E=(-1,1,1)所以(1-m)x+y-2z=0-x+y+z=0,令x=3,得y=m+1,z=2-m,于是,面DFE的一个法向量为n2=(3,m+1,2-m),所以cos(n1,n2〉=)2设面BCC与面DFE所成的二面角为6,则in6=√1-m(n1,m2),故当m=时,面BBCC与面DFE所成的二面角的正弦值最小为,即当B1D=时,面BB1C1C与面DFE所成的二面角的正弦值最小.【解題关键】本题求解关键是建立恰当的空间直角坐标系,确定相关点的坐标,再利用空间向量进行运算

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