山东省烟台市部分区县2022届高三下学期三模数学试题（PDF版含答案）

2022 年高考适应性练习 数学参考答案 一、选择题 D A C B C A D C 二、选择题 9.ABC 10.ACD 11. BD 12.AD 三、填空题 3 π 3 1 13. 14.2160 15. + 16. 4 12 2 6 四、解答题 17.解：若选① (2b c) cos A = a cosC ，（1）由正弦定理可得， 2sin B cos A sin C cos A = sin AcosC ， ················································ 1分 即 2sin B cos A = sin(A+C) = sin B . ························································ 2分 因为 B∈ (0,π )，所以sin B ≠ 0 ， cos A 1所以 = ， ········································································ 3分 2 因为 A∈ (0,π ) A π，所以 = . ···························································· 4分 3 1 1 故 ABC 的面积为 bc sin A = ×3×4 3× = 3 3 ； ································ 6分 2 2 2 若选②a sin B = 3bcos A，（1）由正弦定理可得， sin Asin B 3 sin B cos A = 0 . ··················································· 1分 因为 B∈ (0,π )，所以sin B ≠ 0 ， 则有sin A 3 cos A = 2sin(A π ) = 0， ············································ 3分 3 因为 A π∈ (0,π )，所以 A = . ················································· 4分 3 高三数学参考答案（第 1 页，共 8 页） ABC 1故 的面积为 bc sin A 1 3= ×3×4× = 3 3 ； ··································· 6分 2 2 2 若选③a cosC + 3ac sin A = b + c ，（1）由正弦定理可得， sin AcosC + 3 sin Asin C = sin B + sin C ， ··············································· 1分 3 sin Asin C = cos Asin C + sin C ， ······································· 2分 因为C∈ (0,π ) ，所以sin C ≠ 0， 所以 3 sin A cos A =1，可得sin(A π 1 ) = , ·································· 3分 6 2 因为 A∈ (0,π ) π，所以 A = . ················································ 4分 3 ABC 1故 的面积为 bc sin A 1 3= ×3×4× = 3 3 ； ·································· 6分 2 2 2 （2）在 ABC 2中，由余弦定理可得，a = b2 + c2 2bc cos A = 9+16 2 1×3×4× =13，故a = 13 . ········································· 8分 2 在 ABD 中，2AD×BD cos∠ADB = AD2 + BD2 16， 在 ACD中，2AD×CD cos∠ADC = AD2 +CD2 9， BD CD 13又 = = ,∠ADC +∠ADB = π ,两式相加可得， 2 AD2 37= AD 37，即 = . ··············································· 10分 4 2 在 ACD中，由余弦定理可得， 高三数学参考答案（第 2 页，共 8 页） 37 13 cos ADC AD 2 +CD2 AC 2 + 9 ∠ = = 4 4 7 481= . ·················· ... ...