《2025年小题狂做高中数学必修第二册人教版》

1. 在$\triangle ABC$中,角$A$,$B$,$C$所对的边分别为$a$,$b$,$c$,且$c = 4\sqrt{2}$,$B = 45^{\circ}$,面积$S = 2$,则$b$ =(B)

A.$\frac{\sqrt{113}}{2}$
B.$5$
C.$\sqrt{41}$
D.$25$

答案： 1. B 因为$c = 4\sqrt{2}$，$B = 45^{\circ}$，所以面积$S = \frac{1}{2}ac\sin B = \frac{1}{2} × 4\sqrt{2} × \frac{\sqrt{2}}{2}a = 2$，解得$a = 1$。由余弦定理得$b^{2} = a^{2} + c^{2} - 2ac\cos B = 1 + 32 - 2 × 1 × 4\sqrt{2} × \frac{\sqrt{2}}{2} = 25$，所以$b = 5$。

2. 教材变式 已知$a$,$b$,$c$分别为$\triangle ABC$的三个内角$A$,$B$,$C$的对边. 若$\frac{\sin C - \sin B}{\sin A} = \frac{a - b}{c + b}$,则$C$ =(C)

A.$\frac{\pi}{6}$
B.$\frac{\pi}{4}$
C.$\frac{\pi}{3}$
D.$\frac{5\pi}{12}$

答案： 2. C 根据正弦定理及$\frac{\sin C - \sin B}{\sin A} = \frac{a - b}{c + b}$，得$\frac{c - b}{a} = \frac{a - b}{c + b}$，化简得$a^{2} + b^{2} - c^{2} = ab$，所以$\cos C = \frac{a^{2} + b^{2} - c^{2}}{2ab} = \frac{ab}{2ab} = \frac{1}{2}$。又$C \in (0,\pi)$，所以$C = \frac{\pi}{3}$。教材链接 人教A版必修二习题6.4第22题改编

3. 在$\triangle ABC$中,角$A$,$B$,$C$所对的边分别为$a$,$b$,$c$,已知$A = 60^{\circ}$,$b = 2$,$S_{\triangle ABC} = 2\sqrt{3}$,则$\frac{a + b + c}{\sin A + \sin B + \sin C}$ =(D)

A.$\sqrt{3}$
B.$2$
C.$2\sqrt{3}$
D.$4$

答案： 3. D 由$S_{\triangle ABC} = \frac{1}{2}bc\sin A = 2\sqrt{3}$，$A = 60^{\circ}$，$b = 2$，可得$c = 4$。由余弦定理知$a = \sqrt{b^{2} + c^{2} - 2bc\cos A} = \sqrt{4 + 16 - 2 × 2 × 4 × \frac{1}{2}} = 2\sqrt{3}$，所以由正弦定理可知$\frac{a + b + c}{\sin A + \sin B + \sin C} = \frac{a}{\sin A} = \frac{2\sqrt{3}}{\frac{\sqrt{3}}{2}} = 4$。

4. $\triangle ABC$的内角$A$,$B$,$C$的对边分别为$a$,$b$,$c$,已知$a\sin A - b\sin B = 4c\sin C$,$\cos A = -\frac{1}{4}$,则$\frac{b}{c}$ =(A)

A.$6$
B.$5$
C.$4$
D.$3$

答案： 4. A 因为$a\sin A - b\sin B = 4c\sin C$，所以由正弦定理得$a^{2} - b^{2} = 4c^{2}$，即$a^{2} = 4c^{2} + b^{2}$。由余弦定理得$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc} = \frac{b^{2} + c^{2} - (4c^{2} + b^{2})}{2bc} = \frac{- 3c^{2}}{2bc} = - \frac{1}{6}$，所以$\frac{b}{c} = 6$。

5. 在$\triangle ABC$中,已知$2a\sin A = (2b - c)\sin B + (2c - b)\sin C$,$a = 2$,则$\triangle ABC$周长的最大值为(B)

A.$4$
B.$6$
C.$\frac{4\sqrt{3}}{3}$
D.$\frac{4\sqrt{3}}{3} + 2$

答案： 5. B 在$\triangle ABC$中，由$2a\sin A = (2b - c)\sin B + (2c - b)\sin C$及正弦定理，得$2a^{2} = (2b - c)b + (2c - b)c$，整理得$a^{2} = b^{2} + c^{2} - bc$。而$a = 2$，因此$4 = (b + c)^{2} - 3bc \geqslant (b + c)^{2} - 3(\frac{b + c}{2})^{2} = \frac{1}{4}(b + c)^{2}$，解得$b + c \leqslant 4$，当且仅当$b = c = 2$时，取等号，所以$\triangle ABC$周长的最大值为$6$。

6. 在$\triangle ABC$中,角$A$,$B$,$C$所对的边分别为$a$,$b$,$c$,已知$c(\sin A - \sin C) = (a - b)(\sin A + \sin B)$,若$\triangle ABC$的面积为$\frac{\sqrt{3}}{4}$,周长为$3b$,则边$AC$上的高为(B)


A.$\frac{\sqrt{3}}{3}$
B.$\frac{\sqrt{3}}{2}$
C.$\sqrt{3}$
D.$2\sqrt{3}$

答案： 6. B 在$\triangle ABC$中，由正弦定理及$c(\sin A - \sin C) = (a - b)(\sin A + \sin B)$，得$c(a - c) = (a - b)(a + b)$，即$a^{2} + c^{2} - b^{2} = ac$。由余弦定理得$\cos B = \frac{a^{2} + c^{2} - b^{2}}{2ac} = \frac{1}{2}$，则$\sin B = \frac{\sqrt{3}}{2}$。由$\triangle ABC$的面积为$\frac{\sqrt{3}}{4}$，得$\frac{1}{2}ac\sin B = \frac{\sqrt{3}}{4}ac = \frac{\sqrt{3}}{4}$，解得$ac = 1$。由$a^{2} + c^{2} - b^{2} = ac$，得$(a + c)^{2} - b^{2} = 3ac$，又周长$a + b + c = 3b$，即$a + c = 2b$，故$4b^{2} - b^{2} = 3ac = 3$，因此$b = 1$。令边$AC$上的高为$h$，则$\frac{1}{2}bh = \frac{\sqrt{3}}{4}$，所以$h = \frac{\sqrt{3}}{2}$。

7. 记$\triangle ABC$的内角$A$,$B$,$C$所对的边分别是$a$,$b$,$c$. 若$b ＞ c$,$\cos B = \frac{\sqrt{6}}{4}$,$a = \frac{\sqrt{6}}{2}b$,则(BD)

A.$A = \frac{\sqrt{6}}{2}B$
B.$A = 2B$
C.$b = \sqrt{3}c$
D.$b = 2c$

答案： 7. BD 因为$\cos B = \frac{\sqrt{6}}{4}$，$B$是三角形内角，所以$\sin B = \sqrt{1 - \cos^{2}B} = \sqrt{1 - (\frac{\sqrt{6}}{4})^{2}} = \frac{\sqrt{10}}{4}$。又$a = \frac{\sqrt{6}}{2}b$，由正弦定理可得$\sin A = \frac{a\sin B}{b} = \frac{\frac{\sqrt{6}}{2}b × \frac{\sqrt{10}}{4}}{b} = \frac{\sqrt{15}}{4}$。又因为$\sin 2B = 2\sin B\cos B = 2 × \frac{\sqrt{10}}{4} × \frac{\sqrt{6}}{4} = \frac{\sqrt{15}}{4}$，所以$\sin A = \sin 2B$，所以$A = 2B$或$A + 2B = \pi$。由$b ＞ c$知$B ＞ C$，所以若$A + 2B = \pi$，又$A + B + C = \pi$，则$B = C$，这与$B ＞ C$矛盾，所以$A = 2B$，故A错误，B正确。由余弦定理可得$b^{2} = (\frac{\sqrt{6}}{2}b)^{2} + c^{2} - 2 × \frac{\sqrt{6}}{2}b × c × \frac{\sqrt{6}}{4} = \frac{3}{2}b^{2} + c^{2} - \frac{3}{2}bc$，即$b^{2} - 3bc + 2c^{2} = 0$，解得$b = c$（舍）或$b = 2c$，故C错误，D正确。