云南省昆明市2023~2024学年高二期末质量检测试题(数学)

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云南省昆明市2023~2024学年高二期末质量检测试题(数学)试卷答案

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分析(1)已知等式利用正弦定理化简,再利用两角和与差的正弦函数公式及诱导公式变形,求出tanB的值,由B为三角形的内角,利用特殊角的三角函数值即可求出B的度数;
(2)利用已知及三角形的面积公式可求ac=4$\sqrt{2}$,利用余弦定理及平方和公式整理可得:a+c=2$\sqrt{2}+2$,两式联立即可求得a,c的值.

解答解:(1)由已知及正弦定理得:sinA=sinBcosC+sinBsinC①,
∵sinA=sin(B+C)=sinBcosC+cosBsinC②,
∴sinB=cosB,即tanB=1,
∵B为三角形的内角,
∴B=$\frac{π}{4}$;
(2)∵S△ABC=$\frac{1}{2}$acsinB=$\frac{\sqrt{2}}{4}$ac=2,
∴解得:ac=4$\sqrt{2}$.①
∵由已知及余弦定理得:4=a2+c2-2accos$\frac{π}{4}$=a2+c2-$\sqrt{2}$ac=(a+c)2-2ac-$\sqrt{2}$ac=(a+c)2-8$\sqrt{2}$-8,
∴解得:(a+c)2=12+8$\sqrt{2}$,可得:a+c=2$\sqrt{2}+2$,②
∴由②可解得:a=2$\sqrt{2}+2$,代入①,整理可得:c2-(2$\sqrt{2}$+2)c+4$\sqrt{2}$=0.
∴解得:$\left\{\begin{array}{l}{c=2\sqrt{2}}\\{a=2}\end{array}\right.$或$\left\{\begin{array}{l}{c=2}\\{a=2\sqrt{2}}\end{array}\right.$.

点评此题考查了正弦、余弦定理,三角形的面积公式,两角和与差的正弦函数公式,以及平方和公式的运用,熟练掌握定理及公式是解本题的关键,属于中档题.