《2025年小题狂做高中数学必修第二册人教版》

9. [2025广东湛江期末]在锐角三角形$ABC$中，角$A,B,C$对应的边分别为$a,b,c$，且$c + b = 2a\cos B$，则下列结论正确的是(ACD)

A.$A = 2B$
B.角$B$的取值范围是$(0,\frac{\pi}{6})$
C.若$\angle BAC$的平分线交$BC$于点$D$，$AD = 2,\sin B = \frac{3}{5}$，则$\frac{1}{b} + \frac{1}{c} = \frac{4}{5}$
D.$\frac{c}{a}$的取值范围是$(\frac{\sqrt{2}}{2},\frac{2\sqrt{3}}{3})$

答案：
9.ACD 由正弦定理及$c + b = 2a\cos B$，得$\sin C + \sin B = \sin (A + B) + \sin B = \sin A\cos B + \sin B\cos A + \sin B = 2\sin A\cos B$，所以$\sin B = \sin A\cos B - \cos A\sin B = \sin (A - B)$。又$0 ＜ A ＜ \frac{\pi}{2}$，$0 ＜ B ＜ \frac{\pi}{2}$，所以$B = A - B$，即$A = 2B$，故A正确。由上知$\begin{cases} 0 ＜ B ＜ \frac{\pi}{2}, \\ 0 ＜ 2B ＜ \frac{\pi}{2}, \\ 0 ＜ \pi - 3B ＜ \frac{\pi}{2}, \end{cases}$可得$\frac{\pi}{6} ＜ B ＜ \frac{\pi}{4}$，故B错误。如图，设$B = \angle BAD = \angle CAD = \theta$，则$\angle ADC = 2\theta$，$\angle ACD = \pi - 3\theta$。在$\triangle ABD$中，由$AD = 2$及正弦定理，得$\frac{c}{\sin (\pi - 2\theta)} = \frac{2}{\sin \theta}$，所以$c = 4\cos \theta$。在$\triangle ACD$中，由正弦定理得$\frac{b}{\sin 2\theta} = \frac{2}{\sin (\pi - 3\theta)}$，则$b = \frac{2\sin 2\theta}{\sin 3\theta}$，所以$\frac{1}{b} + \frac{1}{c} = \frac{\sin 3\theta}{2\sin 2\theta} + \frac{1}{4\cos \theta} = \frac{\sin \theta\cos 2\theta + \cos \theta\sin 2\theta}{2\sin 2\theta} + \frac{1}{4\cos \theta} = \frac{2\cos^2\theta - 1 + \cos \theta}{2\cos \theta} + \frac{1}{4\cos \theta} = \frac{2\cos^2\theta - 1 + \cos \theta + \frac{1}{2}}{4\cos \theta} = \cos \theta$。而$\sin \theta = \sin B = \frac{3}{5}$，且$\frac{\pi}{6} ＜ \theta ＜ \frac{\pi}{4}$，则$\cos \theta = \frac{4}{5}$，所以$\frac{1}{b} + \frac{1}{c} = \frac{4}{5}$，故C正确。由正弦定理得$\frac{c}{a} = \frac{\sin C}{\sin A} = \frac{\sin (\pi - 3B)}{\sin 2B} = \frac{\sin 3B}{\sin 2B} = \frac{\sin B\cos 2B + \cos B\sin 2B}{\sin 2B} = \frac{2\cos^2B - 1 + \cos B}{2\cos B} = 2\cos B - \frac{1}{2\cos B}$。设$t = \cos B \in (\frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2})$，则$y = 2t - \frac{1}{2t}$在$(\frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2})$上单调递增，其值域为$(\frac{\sqrt{2}}{2}, \frac{2\sqrt{3}}{3})$，故D正确。

10. [2025江苏南通期末]在$\triangle ABC$中，$AB = \sqrt{3},AC = 2\sqrt{3},\overrightarrow{CD} = 2\overrightarrow{DB}$，且$\overrightarrow{AD}·\overrightarrow{BC} = 1$，则$\angle BAC =$$\frac{2\pi}{3}$.

答案： 10.$\frac{2\pi}{3}$ 由$\overrightarrow{CD} = 2\overrightarrow{DB}$，知$\overrightarrow{BC} = 3\overrightarrow{BD}$，故$\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BD} = \overrightarrow{AB} + \frac{1}{3}\overrightarrow{BC} = \overrightarrow{AB} + \frac{1}{3}(\overrightarrow{AC} - \overrightarrow{AB}) = \frac{2}{3}\overrightarrow{AB} + \frac{1}{3}\overrightarrow{AC}$。又$\overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB}$，所以$\overrightarrow{AD} · \overrightarrow{BC} = \frac{1}{3}(2\overrightarrow{AB} + \overrightarrow{AC}) · (\overrightarrow{AC} - \overrightarrow{AB}) = \frac{1}{3}(\overrightarrow{AB} · \overrightarrow{AC} - 2\overrightarrow{AB}^2 + \overrightarrow{AC}^2) = 1$，则$\overrightarrow{AB} · \overrightarrow{AC} - 2\overrightarrow{AB}^2 + \overrightarrow{AC}^2 = \sqrt{3} × 2\sqrt{3} × \cos \angle BAC - 2 × 3 + 12 = 3$，得$\cos \angle BAC = -\frac{1}{2}$。由于$0 ＜ \angle BAC ＜ \pi$，故$\angle BAC = \frac{2\pi}{3}$。

11. [2025山东滨州期末]在锐角三角形$ABC$中，角$A,B,C$的对边分别为$a,b,c$.若$(a + b)(\sin A - \sin B) = (c - b)\sin C,a = 2\sqrt{3}$，则$\triangle ABC$周长的取值范围是$(6 + 2\sqrt{3}, 6\sqrt{3}]$.

答案： 11.$(6 + 2\sqrt{3}, 6\sqrt{3}]$ 在$\triangle ABC$中，由$(a + b)(\sin A - \sin B) = (c - b)\sin C$及正弦定理，得$(a + b)(a - b) = (c - b)c$，即$a^2 = b^2 + c^2 - bc$。由余弦定理得$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{bc}{2bc} = \frac{1}{2}$。而$0 ＜ A ＜ \pi$，得$A = \frac{\pi}{3}$。由$\triangle ABC$是锐角三角形，得$\begin{cases} 0 ＜ C ＜ \frac{\pi}{2}, \\ 0 ＜ \frac{2\pi}{3} - C ＜ \frac{\pi}{2}, \end{cases}$则$\frac{\pi}{6} ＜ C ＜ \frac{\pi}{2}$。由正弦定理$\frac{b}{\sin B} = \frac{c}{\sin C} = \frac{a}{\sin A} = \frac{2\sqrt{3}}{\frac{\sqrt{3}}{2}} = 4$，得$b + c = 4\sin B + 4\sin C = 4\sin (\frac{2\pi}{3} - C) + 4\sin C = 4(\frac{\sqrt{3}}{2}\cos C + \frac{1}{2}\sin C) + 4\sin C = 4\sqrt{3}\sin (C + \frac{\pi}{6})$。因为$\frac{\pi}{6} ＜ C ＜ \frac{\pi}{2}$，所以$\frac{\pi}{3} ＜ C + \frac{\pi}{6} ＜ \frac{2\pi}{3}$，所以$4\sqrt{3}\sin (C + \frac{\pi}{6}) \in (6, 4\sqrt{3}]$，所以$a + b + c \in (6 + 2\sqrt{3}, 6\sqrt{3}]$，所以$\triangle ABC$的周长的取值范围是$(6 + 2\sqrt{3}, 6\sqrt{3}]$。

12. [2025江西上饶期末]已知在$\triangle ABC$中，$AB = 4,AC = 2,|\lambda\overrightarrow{AB} + (2 - 2\lambda)\overrightarrow{AC}|(\lambda\in\mathbf{R})$的最小值为$2\sqrt{3}$.若$M$为边$AB$上任意一点，$N$为边$BC$的中点，则$\overrightarrow{NM}·\overrightarrow{CM}$的最小值是$\frac{15}{16}$.

答案：
12.$\frac{15}{16}$ 如图，延长AC，使得$\overrightarrow{AD} = 2\overrightarrow{AC}$，令$\overrightarrow{AG} = \lambda\overrightarrow{AB} + (2 - 2\lambda)\overrightarrow{AC} = \lambda\overrightarrow{AB} + (1 - \lambda)\overrightarrow{AD} (\lambda \in \mathbf{R})$，所以B，G，D三点共线。当$AG \perp BD$时，$|\overrightarrow{AG}|$取最小值$2\sqrt{3}$，过A作$AE \perp BD$于E，所以在$Rt \triangle ABE$中，$AE = 2\sqrt{3}$，$AB = 4$，$\sin \angle ABE = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$，得$\angle ABD = 60^{\circ}$。又因为$AB = AD = 4$，所以$\triangle ABD$是等边三角形，所以$BC \perp AD$。在$Rt \triangle ABC$中，$BC = 2\sqrt{3}$，取NC中点为H，连接MH，则$\overrightarrow{MN} + \overrightarrow{MC} = 2\overrightarrow{MH}$，$\overrightarrow{MN} - \overrightarrow{MC} = 2\overrightarrow{CH}$，将两式平方，并作差，化简得$\overrightarrow{MN} · \overrightarrow{MC} = \overrightarrow{MH}^2 - \overrightarrow{CH}^2 = |\overrightarrow{MH}|^2 - \frac{|\overrightarrow{BC}|^2}{4} = |\overrightarrow{MH}|^2 - \frac{3}{4}$。当$MH \perp AB$时，$MH$取最小值。此时在$Rt \triangle BMH$中，$MH = \frac{1}{2}BH = \frac{1}{2} × \frac{3}{4}BC = \frac{3\sqrt{3}}{4}$，所以$(\overrightarrow{NM} · \overrightarrow{CM})_{\min} = (\frac{3\sqrt{3}}{4})^2 - \frac{3}{4} = \frac{15}{16}$。

13. [2025山东滨州期末]如图，已知$a,b,c$分别为$\triangle ABC$三个内角$A,B,C$的对边，且$a\cos C + \sqrt{3}a\sin C = b$.
(1)求$A$.
(2)已知$D$为边$AB$上的一点，且$BC\perp CD$.
①若$AC = \sqrt{3},AD = 1$，求$AB$；
②求$\frac{AD}{DB}$的取值范围.

答案： 13.解：
(1)在$\triangle ABC$中，由$a\cos C + \sqrt{3}a\sin C = b$及正弦定理，得$\sin A\cos C + \sqrt{3}\sin A\sin C = \sin B = \sin (A + C) = \sin A\cos C + \cos A\sin C$，整理得$\sqrt{3}\sin A\sin C = \cos A\sin C$。而$\sin C ＞ 0$，则$\tan A = \frac{\sqrt{3}}{3}$。又$0 ＜ A ＜ \pi$，所以$A = \frac{\pi}{6}$。
(2)① 在$\triangle ADC$中，由余弦定理，得$CD = \sqrt{AC^2 + AD^2 - 2AC · AD\cos A} = \sqrt{3 + 1 - 2 × \sqrt{3} × 1 × \frac{\sqrt{3}}{2}} = 1$，则$\angle ACD = A = \frac{\pi}{6}$。而$BC \perp CD$，则$\angle BCA = \frac{2\pi}{3}$，$B = \frac{\pi}{6} - A$，所以$AB = 2AC\cos A = 2\sqrt{3} × \frac{\sqrt{3}}{2} = 3$。
②在$\triangle ADC$中，由正弦定理$\frac{CD}{\sin A} = \frac{AD}{\sin \angle ACD}$，得$CD = \frac{AD}{2\sin \angle ACD}$。在$Rt \triangle BCD$中，$CD = BD\sin B$，则$\frac{AD}{2\sin \angle ACD} = BD\sin B$，即$\frac{AD}{BD} = 2\sin \angle ACD · \sin B = \sqrt{3}\sin B\cos B - \sin^2B = \frac{\sqrt{3}}{2}\sin 2B - \frac{1}{2}\cos 2B - \frac{1}{2} = \sin(2B + \frac{\pi}{6}) - \frac{1}{2}$。由$0 ＜ B ＜ \frac{\pi}{3}$，得$\frac{\pi}{6} ＜ 2B + \frac{\pi}{6} ＜ \frac{5\pi}{6}$，于是$\frac{1}{2} ＜ \sin (2B + \frac{\pi}{6}) \leq 1$，则$\frac{AD}{BD} \in (0, \frac{1}{2}]$，所以$\frac{AD}{BD}$的取值范围是$(0, \frac{1}{2}]$。