0  114307  114315  114321  114325  114331  114333  114337  114343  114345  114351  114357  114361  114363  114367  114373  114375  114381  114385  114387  114391  114393  114397  114399  114401  114402  114403  114405  114406  114407  114409  114411  114415  114417  114421  114423  114427  114433  114435  114441  114445  114447  114451  114457  114463  114465  114471  114475  114477  114483  114487  114493  114501  447090 

26. China is famous _________ the Great Wall and also ________ one of the largest countries with the largest population.

A. as, for           B. for, for       C. as, as          D. for, as 

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25. —What ________  when the gang broke in?

— I ________  the TV and was going to bed.

A.were you doing; was turning off    B.did you do; has turned off

C.had you done; was turning off     D.were you doing; had turned off

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24. Both my patents are ________ now. They are both ________teachers.

A. retiring; retired    B. retired; retired C. retired; retiring     D. retiring; retiring

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23. He made up his mind to _________ a new record in the 100-meter race.

   A. put down   B. set down    C. put up        D. set up

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22. She ______ Japanese when she was in Japan. Now she can speak it freely. A. picked out        B. made out      C. made up      D. picked up

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第二部分 英语知识运用

第一节 单项填空(共15小题;每小题1分, 满分15分)

   从A、B、C、D四个选项中, 选出可以填入空白处的最佳选项, 并在答题卡上将该项涂黑。

21.  ______, they were not afraid to face the flood.

   A. Although surprised B. Because surprised C. If surprising     D.As long as surprising

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22.(本小题满分12分)设角ABC是△ABC的三个内角,已知向量m=(sinA+sinC,sinB-sinA),n=(sinA-sinC,sinB),且mn.

(1)求角C的大小;

(2)若向量s=(0,-1),t=(cosA,2cos2),求|st|的取值范围.

解:(1)由题意得m·n=(sin2A-sin2C)+(sin2B-sinAsinB)=0,

即sin2C=sin2A+sin2B-sinAsinB,设abc为内角ABC所对的边长,由正弦定理得c2a2b2ab

再由余弦定理得cosC==,∵0<C<π,

C=.

(2)∵st=(cosA,2cos2-1)=(cosA,cosB),

∴|st|2=cos2A+cos2B

=cos2A+cos2(-A)

=+

=cos2A-sin2A+1

=-sin(2A-)+1,

∵0<A<,∴-<2A-<,

∴-<sin(2A-)≤1,

∴≤|st|2<,∴≤|st|<.

 

 

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21.(本小题满分12分)[2014·长沙一模]风景秀美的凤凰湖畔有四棵高大的银杏树,记作ABPQ,欲测量PQ两棵树和AP两棵树之间的距离,但湖岸部分地方围有铁丝网不能靠近,现可测得AB两点间的距离为100 m,如图,同时也能测量出∠PAB=75°,∠QAB=45°,∠PBA=60°,∠QBA=90°,则PQ两棵树和AP两棵树之间的距离各为多少?

解:在△PAB中,∠APB=180°-(75°+60°)=45°,

由正弦定理得=,解得AP=50.

在△QAB中,∠ABQ=90°,∴AQ=100.

又∠PAQ=75°-45°=30°,

由余弦定理得PQ2AP2AQ2-2AP·AQ·cos∠PAQ=(50)2+(100)2-2×50×100×cos30°=5000,

PQ==50.

PQ两棵树之间的距离为50 m,AP两棵树之间的距离为50 m.

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