发布时间 : 星期一 文章(完整word版)高一必修五解三角形复习题及答案更新完毕开始阅读f0a19e317dd5360cba1aa8114431b90d6c858910
∵0?A??,??661?2sinBcosB22⑵. 由题知,整理得sinB?sinBcosB?2cosB?0, ??322cosB?sinB?A???5????,∴A??,A?.
3666cosB?0,∴tan2B?tanB?2?0,∴tanB?2或tanB??1,
而tanB??1使cosB?sinB?0,舍去,∴tanB?2. 18. 【解】⑴.由余弦定理及已知条件得,a?b?ab?4, 又因为△ABC的面积等于3,所以22221absinC?3,得ab?4. 2?a2?b2?ab?4,联立方程组?解得a?2,b?2.
?ab?4,⑵.由题意得sin(B?A)?sin(B?A)?4sinAcosA, 即sinBcosA?2sinAcosA,
?当cosA?0时,A?,△ABC是直角三角形;
2当cosA?0时,得sinB?2sinA?2sin(B?C)?2sinBcosC?2cosBsinC,
??C?代入上式得sinB?sinB?3cosB,故cosB?0,B?,
32△ABC是直角三角形.
19.【解】⑴.由a?2bsinA,根据正弦定理得sinA?2sinBsinA,所以sinB?1, 2π. 6???⑵.cosA?sinC?cosA?sin????A?
???????cosA?sin??A?
?6?13?cosA?cosA?sinA
22????3sin?A??.
3??由△ABC为锐角三角形知,
????0?A?,0?C???A?B?,解得?A?
22322?????A??, 3361???33??3?所以sin?A???,?3sin?A????3,
2?3?223?2?由△ABC为锐角三角形得B?故cosA?sinC的取值范围为??33?. ??2,?2??20. 【解】如图,连结A1B1,由已知A2B2?102,A1A2?302?20?102, 60?A1A2?A2B1,
ooo又∠A1A2B2?180?120?60,?△A1A2B2是等边三角形,
?A1B2?A1A2?102,
ooo由已知,A1B1?20,∠B1A1B2?105?60?45,
在△A1B2B1中,由余弦定理,
22B1B2?A1B12?A1B2?2A1B2?A1B2?cos45o
?202?(102)2?2?20?102?2 2?200.
?B1B2?102.
故乙船的速度的大小为
102?60?302(海里/小时). 20oo21. 【选做题】【解法一】如图,在等腰△ABC中,?BAC?36,?ABC??ACB?72,
?ABC的角平分线交AC于D,设BC=1,AB=x,利用此图来求cos36o.
易知△ABC与△BCD相似,故
5?1ABBCx1,即?,解得x?. ?2BCCD1x?1ox2?x2?125?1?△ABC中,由余弦定理,cos36?; 22x4【解法二(用二倍角公式构造方程,解方程)】
cos144?2cos72?1?2?2cos36?1??1,即?cos36o?2?2cos236o?1??1,
o2o2o22设cos36?x,则?x?22x?1?1,可化为8x?8x?x?1?0,
o?2?242?x?1??8x3?8x2?1??0,因x?1?0,故8x3?8x2?1?0, ?2x?1??4x2?2x?1??0,因x?x?12,故4x?2x?1?0, 25?1?5?15?1o?0舍去)(x?,故cos36?.
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