中,ACBABABC??sinsin∴)26(1050015sin21210000????BC(8分)∵ADCD?,高二理科数学第页(共4页)3∴BDBC????=)26(10500?22?=)13(10500?=)17.1(10500?=7350(10分)山顶的海拔高度=10000-7350=2650(米)=2.65千米(12分)21.解:(1)由2223abcab???,得222322abcab???.(2分)由余弦定理知3cos2C?,∴6C??.(4分)(2)∵21cos2cossin12sin[()]122AAmBAC???????????cossin()cossin()6AACAA???????31cossincoscossincossincos6622AAAAAA????????13cossincoscossinsincos()22333AAAAA?????????(8分)∵203A???∴33A??????.(10分)∴11cos()32A?????,即m的取值范围是1[1,)2?.(12分)22、(1)由已知得111ba??,且1112nnnaann????(2分)高二理科数学第页(共4页)4即112nnnbb???从而2112bb??32212bb??……111(2)2nnnbbn?????(4分)于是121111......222nnbb??????=112(2)2nn???又11b?故所求的通项公式1122nnb???(6分)(2)由(I)知111(2)222nnnnann??????,?nS=11(2)2nkkkk????111(2)2nnkkkkk???????(9分)而1(2)(1)nkknn????,又112nkkk???是一个典型的错位相减法模型,易得1112422nknkkn????????nS=(1)nn?1242nn????(12分)
全文预览