《2024年高职高考全真模拟试卷辽海出版社高中数学全一册人教版》

22. a，b，c分别是△ABC的内角A，B，C的对边，S是△ABC的面积，若a = 4，b = 5，S = 5$\sqrt{3}$，求c边长.

答案： 解：由$S=\frac{1}{2}ab\sin C=\frac{1}{2}\cdot4\cdot5\cdot\sin C = 5\sqrt{3}$得$\sin C=\frac{\sqrt{3}}{2}$，则$\cos C=\frac{1}{2}$或$-\frac{1}{2}$当$\cos C=\frac{1}{2}$时，$c^{2}=a^{2}+b^{2}-2ab\cos C = 4^{2}+5^{2}-2\times4\times5\times\frac{1}{2}=21$，得$c = \sqrt{21}$当$\cos C=-\frac{1}{2}$时，$c^{2}=a^{2}+b^{2}-2ab\cos C = 4^{2}+5^{2}+2\times4\times5\times\frac{1}{2}=61$，得$c = \sqrt{61}$综上所述$c$为$\sqrt{21}$或$\sqrt{61}$

23. 已知等差数列{aₙ}的前n项和为Sₙ，等比数列{bₙ}的前n项和为Tₙ，a₁ = -1，b₁ = 1，a₂ + b₂ = 2.
(1)若a₃ + b₃ = 5，求{bₙ}的通项公式；
(2)若T₃ = 21，求S₃.

答案： 解：
(1)设公差为$d$，公比为$q$由$\begin{cases}a_{2}+b_{2}=(-1 + d)+q = 2\\a_{3}+b_{3}=(-1 + 2d)+q^{2}=5\end{cases}$，得$q = 2$或$q = 0$(舍去)，$d = 1$因此，$\{b_{n}\}$的通项公式为$b_{n}=2^{n - 1}$
(2)$\because T_{3}=b_{1}+b_{2}+b_{3}=1 + q+q^{2}=21$$\therefore q=-5$或$4$又$\because a_{2}+b_{2}=(-1 + d)+q = 2$$\therefore d = 8$或$-1$则$S_{3}=3\times(-1)+\frac{3\times2}{2}\times8 = 21$或$S_{3}=3\times(-1)+\frac{3\times2}{2}\times(-1)=-6$