必修5第一章解三角形测试题及答案

必修五第一章解三角形测试(总分150)一、选择题(每题5分，共50分)1、在△ABC中，a＝3，b＝7，c＝2，那么B等于（）A．30°B．45°C．60°D．120°2、在△ABC中，a＝10，B=60°,C=45°,则c等于（）A．310B．1310C．13D．3103、在△ABC中，a＝32，b＝22，B＝45°，则A等于（）A．30°B．60°C．30°或120°D．30°或150°4、在△ABC中，3AB,1AC,∠A＝30°，则△ABC面积为（）A．23B．43C．23或3D．43或235、在△ABC中，已知bccba222，则角A为（）A．3B．6C．32D．3或326、在△ABC中,面积22()Sabc,则sinA等于（）A．1517B．817C．1315D．13177、已知△ABC中三个内角为A、B、C所对的三边分别为a、b、c，设向量(,)pacb，(,)qbaca.若//pq，则角C的大小为（）A．6B．3C．2D．238、已知锐角三角形的边长分别为1，3，a，则a的范围是（）A．10,8B．10,8C．10,8D．8,109、在△ABC中，已知CBAsincossin2,那么△ABC一定是()A．直角三角形B．等腰三角形C．等腰直角三角形D．正三角形10、在△ABC中,3,13,4ABBCAC,则AC上的高为（）nm3111QPBAA．322B．332C．32D．33二、填空题（每小题5分，共20分）11、在△ABC中，若∠A:∠B:∠C=1:2:3,则cba::12、已知三角形两边长为1和3，第三边的中线长为1，则第三边长为13、若三角形两边长为1和3，第三边上的中线长为1，则三角形的外接圆半径为14、在△ABC中BC=1，3B，当△ABC面积为3时，tanC三、解答题（本大题共小题6小题，共80分）15、（本小题14分）在△ABC中，已知210AB，A＝45°，在BC边的长分别为20，3320，5的情况下，求相应角C。16、（本小题14分）在△ABC中，BC＝a，AC＝b，a，b是方程02322xx的两个根，且1cos2BA。求：(1)角C的度数；(2)AB的长度。17、（本小题12分）在△ABC中，角A、B、C所对的三边分别为a、b、c，22sin3cos,7CCc，又△ABC的面积为332.求：（1）角C大小（2）ab的值18．（本小题12分）在△ABC中，10ba，cosC是方程02322xx的一个根，求△ABC周长的最小值。19．（本小题14分）在△ABC中，若BACBAcoscossinsinsin.(1)判断△ABC的形状；(2)在上述△ABC中，若角C的对边1c，求该三角形内切圆半径的取值范围。20．如图所示，平面上有四点A、B、Q、P，其中A、B为定点，且3AB，P、Q为动点，满足1APPQQB，△APB与△PQB的面积分别为,mn.(1)设30A,求Q(2)求22mn的最大值必修五第一章测试答案一、选择题(每题5分，共50分)12345678910CBCBCBBBBB二、填空题（每小题5分，共20分）11.12.213.114.23三、解答题（本大题共小题6小题，共80分）15、（本小题14分）解：由正弦定理得BCBCAABC10sinsin（1）当BC＝20时，sinC＝21；ABBCCA30C°（2）当BC＝3320时，sinC＝23；ABBCAB45sinC有两解60C或120°（3）当BC＝5时，sinC＝21；C不存在16、（本小题14分）解：（1）21coscoscosBABACC＝120°（2）由题设：322baab120cos2cos222222abbaCBCACBCACAB102322222abbaabba10AB17、（本小题12分）(1)60C(2)5ab18．（本小题12分）解：02322xx21,221xx又Ccos是方程02322xx的一个根21cosC由余弦定理可得：abbaabbac2222212则：7551010022aaac当5a时，c最小且3575c此时3510cba△ABC周长的最小值为351019．（本小题14分）解：（1）由BACBAcoscossinsinsin可得12sin22C0cosC即C＝90°△ABC是以C为直角顶点得直角三角形（2）内切圆半径cbar211sinsin21BA212214sin22A内切圆半径的取值范围是212,020．(1)60Q(2)7820·2007·新疆奎屯[e-mail]特级教师王新敞源头学子小屋本小题主要考查等差数列，等比数列等基础知识，考查基本运算能力·2007·新疆奎屯[e-mail]特级教师分·2007·新疆奎屯[e-mail]特级教师王新敞源头学子小屋（Ｉ）解：由题设得11()2(2)nnnnababn≥，········································1分即12nncc（2n≥）················································································2分易知{}nc是首项为113ab，公差为２的等差数列，···········································3分通项公式为21ncn·2007·新疆奎屯[e-mail]特级教师王新敞源头学子小屋·················································································4分2(321)22nnnSnn············································································6分（II）解：由题设得111()(2)2nnnnababn≥，令nnndab，则11(2)2nnddn≥·2007·新疆奎屯[e-mail]特级教师王新敞源头学子小屋易知{}nd是首项为111ab，公比为12的等比数列，··········································7分通项公式为112nnd·2007·新疆奎屯[e-mail]特级教师王新敞源头学子小屋···················································································8分由12112nnnnnabnab，························································································9分解得1122nnan，··················································································10分3298a··································································································11分3332927788bca··········································································12分12nnTaaa22111122222nn（）（）··························································13分21122nnn·2007·新疆奎屯[e-mail]特级教师王新敞源头学子小屋···················································································14分