《2026年南方凤凰台5A新考案数学二轮提高版》

例 2 在$\triangle ABC$中，内角$A$，$B$，$C$的对边分别为$a$，$b$，$c$，且$b\sin B\tan A = \sqrt{3}a\sin B\cos C + \sqrt{3}b\sin C\cos A$。
（1）求角$A$的大小；
（2）若点$D$在边$BC$上，且$BD = 2DC$，$b = 3$，$AD = 2\sqrt{3}$，求$\triangle ABC$的周长。$\frac{\pi}{3}$

答案：
例2　[解答]
(1)因为$b\sin B\tan A=\sqrt{3}a\sin B\cos C+\sqrt{3}b·\sin C\cos A$,所以$\sin^2B\tan A=\sqrt{3}\sin A\sin B\cos C+\sqrt{3}\sin B·\sin C\cos A$.因为$B\in(0,\pi)$,所以$\sin B\neq0$,所以$\sin B\tan A=\sqrt{3}\sin A\cos C+\sqrt{3}\sin C\cos A=\sqrt{3}\sin(A + C)$.因为$\sin(A + C)=\sin B$,所以$\tan A=\sqrt{3}$.因为$A\in(0,\pi)$,所以$A=\frac{\pi}{3}$.
(2)方法一:如图,设$CD = x$,则在$\triangle ABC$中,$\cos\angle BAC=\frac{9 + c^2 - 9x^2}{6c}=\frac{1}{2}$①.在$\triangle ACD$中,$\cos\angle ADC=\frac{12 + 4x^2 - 9}{4\sqrt{3}x}=\frac{3 + x^2}{4\sqrt{3}x}$.在$\triangle ADB$中,$\cos\angle ADB=\frac{12 + 4x^2 - c^2}{8\sqrt{3}x}$.因为$\cos\angle ADC+\cos\angle ADB = 0$,所以$6x^2 + 18 - c^2 = 0$②.由①②可解得$c = 6$,$x=\sqrt{3}$,所以$\triangle ABC$的周长为$3\sqrt{3}+9$.
方法二:因为$BD = 2DC$,所以$\overrightarrow{AD}=\overrightarrow{AB}+\frac{2}{3}\overrightarrow{BC}=\overrightarrow{AB}+\frac{2}{3}(\overrightarrow{AC}-\overrightarrow{AB})=\frac{1}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}$,所以$\overrightarrow{AD}^2=\frac{1}{9}\overrightarrow{AB}^2+\frac{4}{9}\overrightarrow{AB}·\overrightarrow{AC}+\frac{4}{9}\overrightarrow{AC}^2$,即$12=\frac{1}{9}c^2+\frac{4}{9}× c×3×\frac{1}{2}+\frac{4}{9}×9$,即$c^2 + 6c - 72 = 0$,解得$c = 6$或$c=-12$(舍去).由余弦定理得$a^2=36 + 9 - 2×6×3×\frac{1}{2}=27$,所以$a = 3\sqrt{3}$,所以$\triangle ABC$的周长为$3\sqrt{3}+3 + 6 = 3\sqrt{3}+9$.

变式 2 （2025·张家口三模）在$\triangle ABC$中，内角$A$，$B$，$C$的对边分别为$a$，$b$，$c$，已知$2 - \sin A\sin C = \sin^2(A + C) + \cos^2(A + B) + \cos^2(B + C)$。
（1）求角$B$的大小；
（2）若$\triangle ABC$外接圆的面积为$\pi$，且$a + \sqrt{2} = c$，$a\overrightarrow{AD} = c\overrightarrow{CD}$，求$BD$的长。$\frac{\pi}{3}$$\frac{\sqrt{2}}{2}$

答案： 变式2　[解答]
(1)因为$2-\sin A\sin C=\sin^2(A + C)+\cos^2(A + B)+\cos^2(B + C)$,所以$2-\sin A\sin C=\sin^2B+\cos^2C+\cos^2A$,所以$1-\cos^2C + 1-\cos^2A-\sin A\sin C=\sin^2B$,所以$\sin^2B=\sin^2A+\sin^2C-\sin A\sin C$,由正弦定理可知$b^2=a^2 + c^2-ac$,即$a^2 + c^2 - b^2=ac$,又由余弦定理可知$\cos B=\frac{a^2 + c^2 - b^2}{2ac}=\frac{1}{2}$,又$B\in(0,\pi)$,所以$B=\frac{\pi}{3}$.
(2)由$\triangle ABC$的外接圆面积为$\pi$,得外接圆半径为$1$,由正弦定理得$b = 2\sin B=\sqrt{3}$,由余弦定理及$a+\sqrt{2}=c$,得$3=(\sqrt{2}+a)^2+a^2-a(\sqrt{2}+a)$,化简得$a^2+\sqrt{2}a - 1 = 0$,解得$a=\frac{\sqrt{6}-\sqrt{2}}{2}$(负根舍去),从而$c=\frac{\sqrt{6}+\sqrt{2}}{2}$.因为$\overrightarrow{AD}=\frac{c}{c - a}\overrightarrow{AC}$,$\overrightarrow{BD}=\overrightarrow{BA}+\overrightarrow{AD}=\overrightarrow{BA}+\frac{c}{c - a}(\overrightarrow{BC}-\overrightarrow{BA})=-\frac{a}{c - a}\overrightarrow{BA}+\frac{c}{c - a}\overrightarrow{BC}=-\frac{\sqrt{3}-1}{2}\overrightarrow{BA}+\frac{\sqrt{3}+1}{2}\overrightarrow{BC}$,所以$|\overrightarrow{BD}|^2=(-\frac{\sqrt{3}-1}{2}\overrightarrow{BA}+\frac{\sqrt{3}+1}{2}\overrightarrow{BC})^2=\frac{1}{4}[(\sqrt{3}-1)^2\overrightarrow{BA}^2+(\sqrt{3}+1)^2\overrightarrow{BC}^2-2(\sqrt{3}-1)(\sqrt{3}+1)\overrightarrow{BA}·\overrightarrow{BC}]=\frac{1}{4}[(4 - 2\sqrt{3})(\frac{\sqrt{6}+\sqrt{2}}{2})^2+(4 + 2\sqrt{3})(\frac{\sqrt{6}-\sqrt{2}}{2})^2-2×2×\frac{\sqrt{6}+\sqrt{2}}{2}×\frac{\sqrt{6}-\sqrt{2}}{2}×\frac{1}{2}]=\frac{1}{2}$,故BD的长为$\frac{\sqrt{2}}{2}$.