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2018届高三数学(理)一轮复习课后作业:第三章 三角函数、解三角形 第7节 正弦定理和余弦定理

课时作业 A组 基础对点练 π 1.在△ABC 中,角 A,B,C 的对边分别为 a,b,c,且 b2=a2+bc,A=6,则角 C=( π A.6 π 3π C.6或 4 ) π B.4 π 3π D.4或 4 2 2 2 b2+c2-a2 3 b +c -a 2 解析:在△ABC 中,由余弦定理得 cos A= 2bc ,即 2 = 2bc ,所以 b +c2-a2= 3bc.又 b2=a2+bc,所以 c2+bc= 3bc,即 c=( 3-1)b<b,则 a= b2+a2-c2 2 π 2- 3b,所以 cos C= 2ab = 2 ,解得 C=4.故选 B. 答案:B 2.在△ABC 中,角 A,B,C 所对的边分别为 a,b,c,若 b2+c2-a2= 3bc,且 b = 3a,则下列关系一定不成立的是( A.a=c C.2a=c 解析:由余弦定理,得 cos A= ) B.b=c D.a2+b2=c2 b2+c2-a2 3bc 3 = = .又 b= 3a,由正 2bc 2bc 2 ,则 A=30° 3 弦定理得 sin B= 3sin A= 3sin 30° =2, 所以 B=60° 或 120° .当 B=60° 时, △ABC 为直角三角形,且 2a=c,可知 C,D 成立;当 B=120° 时,C=30° ,所以 A=C, 即 a=c,可知 A 成立,故选 B. 答案:B 3.△ABC 中,角 A,B,C 的对边分别是 a,b,c.已知 b=c,a2=2b2(1-sin A), 则 A=( 3π A. 4 π C.4 ) π B.3 π D.6 解析:先根据正弦定理化边为角,然后根据诱导公式、倍角公式等化简. ∵b=c,∴B=C. π A 又由 A+B+C=π 得 B=2- 2 . 由正弦定理及 a2=2b2(1-sin A)得 sin2A=2sin2B(1-sin A), ?π A? 即 sin2A=2sin2?2- 2 ?(1-sin A), ? ? A 即 sin2A=2cos2 2 (1-sin A), A A A 即 4sin2 2 cos2 2 =2cos2 2 (1-sin A), A? A? 整理得 cos2 2 ?1-sin A-2sin2 2 ?=0, ? ? A 即 cos2 2 (cos A-sin A)=0. A π A ∵0<A<π,∴0< 2 <2,∴cos 2 ≠0, π ∴cos A=sin A.又 0<A<π,∴A= . 4 答案:C 4.(2017· 太原模拟)在△ABC 中,角 A,B,C 所对的边分别为 a,b,c,若 A=60° , b=1,S△ABC= 3,则 c 等于( A.1 C.3 ) B.2 D.4 1 1 3 解析:∵S△ABC=2bcsin A,∴ 3=2×1×c× 2 ,∴c=4. 答案:D c 5.(2017· 武汉调研)在△ABC 中,角 A,B,C 所对的边分别为 a,b,c,若b<cos A, 则△ABC 为( A.钝角三角形 C.锐角三角形 ) B.直角三角形 D.等边三角形 c sin C 解析:根据正弦定理得b=sin B<cos A, 即 sin C<sin Bcos A,∵A+B+C=π,∴sin C=sin(A+B)<sin Bcos A,整理得 sin π Acos B<0,又三角形中 sin A>0,∴cos B<0,2<B<π.∴△ABC 为钝角三角形. 答案:A 2π b 6.(2016· 高考北京卷)在△ABC 中,∠A= 3 ,a= 3c,则 c=________. 解析:先利用余弦定理建立关系式,再将所给条件代入求解即可. 在△ABC 中,∠A= 2π , 3 2π ∴a2=b2+c2-2bccos 3 ,即 a2=b2+c2+bc. ∵a= 3c,∴3c2=b2+c2+bc,∴b2+bc-2c2=0, b ∴(b+2c)(b-c)=0,∴b-c=0,∴b=c,∴ c=1. 答案:1 7.在△ABC 中,角 A,B,C 所对的边分别是 a,b,c,若 b2+c2=2a2,则 cos A 的最小值为________. a2 1 解析:因为 b2+c2=2a2,则由余弦定理可知 a2=2bccos A,所以 cos A=2bc=2 b2+c2 1 2bc 1 1 × ≥ × = (当且仅当 b=c 时等号成立),即 cos A 的最小值为 . 2bc 2 2bc 2 2 1 答案:2 π 8.在△ABC 中,A=3,AC=4,BC=2 3,则△ABC 的面积等于________. 解析:由正弦定理 BC AC 2 3 4 π = ,得 = ,解得 sin B=1,B= ,所以△ABC sin A sin B sin B 2 3 2 1 1 为直角三角形, 所以 AB= AC2-BC2=2, 所以 S△ABC=2AB· BC=2×2×2 3=2 3. 答案:2 3 9.已知△ABC 的三边 a,b,c 所对的角分别为 A,B,C,且 a∶b∶c=7∶5∶3. (1)求 cos A 的值; (2)若△ABC 的面积为 45 3,求△ABC 外接圆半径 R 的大小. 解析:因为 a∶b∶c=7∶5∶3,所以可设 a=7k,b=5k, c=3k(k>0). (1)由余弦定理得 cos A= b2+c2-a2 ?5k?2+?3k?2-?7k?2 1 = =- 2bc 2· 5k· 3k 2. 1 (2)由(1)知 cos A=-2,因为 0<A<π, 3 所以 sin A= 1-cos2A= 2 . 1 又△ABC 的面积为 45 3,所以2bcsin A=45 3, 1 3 即2×5k×3k× 2 =45 3,解得 k=2 3或 k=-2 3(舍去). a 7k 14 3 由正弦定理得sin A=2R,得 2R=sin A= =28,即 R=14. 3 2 10.(2017

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