数学第1章解三角形测试1新人教A版必修5高中数学练习试题

第-1-页共6页第一章解三角形单元测试一、选择题1.A为△ABC的内角，则AAcossin的取值范围是（）A.)2,2(B.)2,2(C.]2,1(D.]2,2[2.在△ABC中，若,900C则三边的比cba等于（）A.2cos2BAB.2cos2BAC.2sin2BAD.2sin2BA3.在△ABC中，若8,3,7cba，则其面积等于（）A.12B.221C.28D.364.在△ABC中，090C，00450A，则下列各式中正确的是（）A.sincosAAB.sincosBAC.sincosABD.sincosBB5.在△ABC中，若)())((cbbcaca，则A（）A.090B.060C.0120D.01506.新疆源头学子小屋特级教师王新敞@[e-mail]王新敞特级教师源头学子小屋新疆在△ABC中，若22tantanbaBA，则△ABC的形状是（）A.直角三角形B.等腰或直角三角形C.不能确定D.等腰三角形二、填空题1.在△ABC中，若,sinsinBA则A一定大于B，对吗？填_________（对或错）2.在△ABC中，若,1coscoscos222CBA则△ABC的形状是______________.3.在△ABC中，∠C是钝角，设,coscos,sinsin,sinBAzBAyCx则zyx,,的大小关系是___________________________.4.在△ABC中，若bca2，则CACACAsinsin31coscoscoscos______.5.在△ABC中，若,tanlgtanlgtanlg2CAB则B的取值范围是_______________.6.在△ABC中，若acb2，则BBCA2coscos)cos(的值是_________.第-2-页共6页三、解答题1.在△ABC中，若)sin()()sin()(2222BAbaBAba，请判断三角形的形状.2.如果△ABC内接于半径为R的圆，且,sin)2()sin(sin222BbaCAR求△ABC的面积的最大值.3.已知△ABC的三边cba且2,2CAbca，求::abc4.在△ABC中，若()()3abcabcac，且tantan33AC，AB边上的高为43，求角,,ABC的大小与边,,abc的长第-3-页共6页第一章解三角形参考答案一、选择题1.Csincos2sin(),4AAA而520,sin()144424AAA2.BsinsinsinsinsinabABABcC2sincos2cos222ABABAB3.D011cos,60,sin6322ABCAASbcA4.D090AB则sincos,sincosABBA，00045,AsincosAA，004590,sincosBBB5.C22222201,,cos,1202acbbcbcabcAA6.B22sincossincossin,,sincossincoscossinsincossinABABAAABBABBABsin2sin2,2222ABABAB或二、填空题1.对,sinsinBA则22ababABRR2.直角三角形21(1cos21cos2)cos()1,2ABAB21(cos2cos2)cos()0,2ABAB2cos()cos()cos()0ABABABcoscoscos0ABC3.zyx,,sincos,sincos,22ABABABBAyz,sinsinsin,,cabCABxyxyz第-4-页共6页4.1sinsin2sin,2sincos4sincos2222ACACACACACBcos2cos,coscos3sinsin222222ACACACAC则221sinsin4sinsin322ACAC1coscoscoscossinsin3ACACAC22(1cos)(1cos)14sinsin22ACAC22222sin2sin4sinsin112222ACAC5.)2,3[2tantantantantan,tantan()tantan1ACBACBACAC2tantantantan()tan1ACBACB3tantantantan2tantan2tanBBACACB3tan3tan,tan0tan33BBBBB6.122,sinsinsin,bacBACBBCA2coscos)cos(2coscossinsincos12sinACACBBcoscossinsincos12sinsinACACBACcoscossinsincos1ACACBcos()cos11ACB三、解答题1.解：22222222sin()sincossin,sin()cossinsinabABaABAabABbABBcossin,sin2sin2,222cossinBAABABABAB或2∴等腰或直角三角形2.解：2sinsin2sinsin(2)sin,RAARCCabB222sinsin(2)sin,2,aAcCabBacabb222222022,cos,4522abcabcabCCab第-5-页共6页2222,2sin2,22,sincRcRCRabRabC22222222,22RRabababab21222sin,24422RSabCab2max212RS另法：122sin2sin2sin244SabCabRARB222sin2sin2sinsin4RARBRAB212[cos()cos()]2RABAB22122[cos()]2222(1)22RABR2max212SR此时AB取得等号3.解：sinsin2sin,2sincos4sincos2222ACACACACACB12147sincos,cos,sin2sincos222424224BACBBBB3,,,24242BBACACBAC33371sinsin()sincoscossin4444ABBB71sinsin()sincoscossin4444CBBB::sin:sin:sinabcABC)77(:7:)77(4.解：22201()()3,,cos,602abcabcacacbacBBtantan33tan(),3,1tantan1tantanACACACAC第-6-页共6页tantan23AC，联合tantan33AC得tan1tan23tan1tan23AACC或，即000075454575AACC或当0075,45AC时，434(326),8(31),8sinbcaA当0045,75AC时，4346,4(31),8sinbcaA∴当00075,60,45ABC时，8,4(326),8(31),abc当00045,60,75ABC时，8,46,4(31)abc.