2022届砺剑·2022相约高考三轮冲刺整合卷(四)4老高考版文科综合答案，知嘛答案网已经编辑汇总了2022届砺剑·2022相约高考三轮冲刺整合卷(四)4老高考版文科综合答案的各科答案和试卷，更多试卷答案请关注知嘛答案。
第二节书面表达【写[文]作思路】根据试题要[章]求,考生需要写一篇[来]记叙文,记述身边一[自]位值得尊敬和爱戴的[知]人。首先考生需要介[嘛]绍自己最尊敬和爱戴[答]的人是谁,简单地介[案]绍一下他(她)。然[网]后,陈述为什么他([文]她)值得尊敬和爱戴[章]。最后,可以对前文[来]进行一个总结,使文[自]章结构严谨。本文的[知]写作重点在"尊敬和爱戴的原因",考生可围绕这个要[嘛]点适当拓展,从性格[答]特点、处事风格等方[案]面展开写作。【范文[网]赏读】The most beloved and respected[文] person around me is my teacher, Ms LiThough she has been teaching English for twenty years, she is still passionat[章]eabout teaching.[来] She is kind and considera[自]te towards us just like our dear motherWe all respect her because she always tries new ways to make her classes livelyand[知] interesti[嘛]ng. Hardworki[答]ng and knowledge[案]able, she is one of the best teachersi[网]n our school. When we have a problem, we will turn to her for help. She alwaystal[文]ks to us patiently[章] and helps us to find a solution.[来] In our eyes, she is not onlyour teacher, but also our best friend. Those are why she deserves our respect【亮[自]点词句】亮点词汇:[知] beloved, respected[嘛], be passionat[答]e about, be considera[案]tetowards[网], turn to sb. for help, in one's eyes亮点句式:[文]Why引导的表语从[章]句( why she deserves our respect)
19.几何法:(1[来])证明:易知当PA[自]=AC,E为PC的[知]中点时,AE⊥PC[嘛];且由C点在以AB[答]为直径的圆上,可得[案]AC⊥BC.………[网]2分另外,PA⊥底[文]面ABC,且BCC[章]面ABC,则PA⊥[来]BC而PAc面PA[自]C,ACC面PAC[知],PA∩AC=A,[嘛]可知BC⊥面PAC[答].…………4分因为[案]AEC面PAC,所[网]以BC⊥AE又PC[文]C面PBC,BCC[章]面PBC,且PC∩[来]BC=C,可知AE[自]⊥面PBC,又PB[知]C平面PBC,故A[嘛]E⊥PB.6分(2[答])如图1,过点E作[案]EF⊥PB交PB于[网]点F,由AE⊥PB[文],EF⊥PB可知∠[章]AFE为二面角C-[来]BP-A的平面角,[自]…8分由余弦定理知[知]∞∠AFE、AE=[嘛],BF、原若设PA[答]=a,则可求得AF[案]=⑤10分AE 1则二面角C-BP[网]-A的大小为60°[文].…12分注:若利[章]用(1)中AE⊥面[来]PBC所得AE⊥E[自]F,即RT△AEF[知]中AF=5a,AE[嘛]=2a,也可求得√[答]3sin∠AFE=[案]图空间向量法:由B[网]C⊥AC,BC⊥P[文]A,知BC⊥面PA[章]C,在平面ABC内[来]过A作垂直AC的直[自]线为x轴,ACAP[知]所在的直线为y轴,[嘛]z轴;即以A为坐标[答]原点,建立如图2的[案]空间直角坐标系,…[网]…2分可设PA=a[文](1)若设BC=b[章],则A(0,0,0[来]),B(b,a,0[自]),C(0,a,0[知]),P(0,0,a[嘛]),…4分因此E([答]0号号)其中A=([案],·号),P=(b[网]m,一a),故A,[文]P=故AE⊥PB6[章]分(2)当PA=A[来]C,E为PC的中点[自]时,AE⊥P℃,则[知]由(1)知AE⊥面[嘛]PBC,故可取面P[答]BC的一个法向量为[案]AE=(0.2,)[网];一……8分(2)[文]当PA=AC,E为[章]PC的中点时,AE[来]⊥℃,则由(1)知[自]AE⊥面PBC,故[知]可取面PBC的一个[嘛]法向量为AE=(0[答]2兰)8分当PA=[案]AC=BC=a时,[网]AP=(0,0,a[文]),AB=(a,a[章],0),图2若设面[来]PAB的法向量为u[自]=(x,y,z),[知]则可取u=(1,-[嘛]1,0)10分AB[答]=0,ar+ay=[案]0.则cos(AE[网],)=由图可知二面[文]角C一BPA为锐角[章],所以二面角C一B[来]P一A的大小为60[自].…12分
