《2026年南方凤凰台5A新考案高中数学二轮基础版》

例 1 （2025·苏锡常镇一模）在△ABC 中，角 A，B，C 所对的边分别为 a，b，c，且 2bcos C = 2a + √{2}c.
（1）求角 B 的大小；
（2）若 b = √{13}，c = 2√{2}，D 为 AC 的中点，求 BD.

答案：
(1)方法一：因为$2b\cos C = 2a + \sqrt{2}c$，所以由余弦定理得$2b · \frac{a^{2} + b^{2} - c^{2}}{2ab} = 2a + \sqrt{2}c$，得$a^{2} + c^{2} - b^{2} = - \sqrt{2}ac$，则$\cos B = \frac{a^{2} + c^{2} - b^{2}}{2ac} = - \frac{\sqrt{2}}{2}$。又因为$B \in (0,\pi)$，所以$B = \frac{3\pi}{4}$。
方法二：因为$2b\cos C = 2a + \sqrt{2}c$，所以由正弦定理得$2\sin B\cos C = 2\sin A + \sqrt{2}\sin C$，即$2\sin B\cos C = 2\sin(B + C) + \sqrt{2}\sin C$，即$2\sin B\cos C = 2\sin B\cos C + 2\cos B\sin C + \sqrt{2}\sin C$，所以$2\cos B\sin C = - \sqrt{2}\sin C$。又因为$\sin C ＞ 0$，所以$\cos B = - \frac{\sqrt{2}}{2}$。又因为$B \in (0,\pi)$，所以$B = \frac{3\pi}{4}$。
(2)方法一：在$\triangle ABC$中，由余弦定理得$(\sqrt{13})^{2} = a^{2} + (2\sqrt{2})^{2} - 2 × a × 2\sqrt{2} × ( - \frac{\sqrt{2}}{2})$，解得$a = 1$，所以$\cos C = \frac{a^{2} + b^{2} - c^{2}}{2ab} = \frac{1 + 13 - 8}{2 × 1 × \sqrt{13}} = \frac{3\sqrt{13}}{13}$。在$\triangle BCD$中，由余弦定理得$BD^{2} = 1 + (\frac{\sqrt{13}}{2})^{2} - 2 × 1 × \frac{\sqrt{13}}{2} × \frac{3\sqrt{13}}{13} = \frac{5}{4}$，得$BD = \frac{\sqrt{5}}{2}$。
方法二：因为$\angle BDC + \angle BDA = \pi$，所以$\cos\angle BDC + \cos\angle BDA = 0$，所以$\frac{BD^{2} + CD^{2} - CB^{2}}{2BD · CD} + \frac{BD^{2} + AD^{2} - AB^{2}}{2BD · AD} = 0$，即$\frac{BD^{2} + (\frac{\sqrt{13}}{2})^{2} - 1^{2}}{2BD · \frac{\sqrt{13}}{2}} + \frac{BD^{2} + (\frac{\sqrt{13}}{2})^{2} - (2\sqrt{2})^{2}}{2BD · \frac{\sqrt{13}}{2}} = 0$，解得$BD = \frac{\sqrt{5}}{2}$。
方法三：在$\triangle ABC$中，由余弦定理得$(\sqrt{13})^{2} = a^{2} + (2\sqrt{2})^{2} - 2 × a × 2\sqrt{2} × ( - \frac{\sqrt{2}}{2})$，解得$a = 1$。因为$\overrightarrow{BD} = \frac{1}{2}(\overrightarrow{BA} + \overrightarrow{BC})$，所以$\overrightarrow{BD}^{2} = \frac{1}{4}(\overrightarrow{BA}^{2} + \overrightarrow{BC}^{2} + 2\overrightarrow{BA} · \overrightarrow{BC}) = \frac{1}{4} × [8 + 1 + 2 × 2\sqrt{2} × 1 × ( - \frac{\sqrt{2}}{2})] = \frac{5}{4}$，所以$BD = \frac{\sqrt{5}}{2}$。

变式 1 （2025·张家口三模）在△ABC 中，内角 A，B，C 的对边分别为 a，b，c，已知 2 - sin A sin C = sin²(A + C) + cos²(A + B) + cos²(B + C).
（1）求角 B 的大小；
（2）若△ABC 的外接圆面积为 π，且 a + √{2} = c，a$\overrightarrow{AD}$ = c$\overrightarrow{CD}$，求 BD 的长.$\frac{\sqrt{2}}{2}$

答案：
(1)因为$2 - \sin A\sin C = \sin^{2}(A + C) + \cos^{2}(A + B) + \cos^{2}(B + C)$，所以$2 - \sin A\sin C = \sin^{2}B + \cos^{2}C + \cos^{2}A$，所以$1 - \cos^{2}C + 1 - \cos^{2}A - \sin A\sin C = \sin^{2}B$，所以$\sin^{2}B = \sin^{2}A + \sin^{2}C - \sin A\sin C$。由正弦定理可知$b^{2} = a^{2} + c^{2} - ac$，即$a^{2} + c^{2} - b^{2} = ac$。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$a\overrightarrow{AD} = c\overrightarrow{CD}$，所以$\overrightarrow{AD} = \frac{c}{a}\overrightarrow{CD} = \frac{c}{c - a}\overrightarrow{AC}$，$\overrightarrow{BD} = \overrightarrow{BA} + \overrightarrow{AD} = \overrightarrow{BA} + \frac{c}{c - a}\overrightarrow{AC} = \overrightarrow{BA} + \frac{c}{c - a}(\overrightarrow{BC} - \overrightarrow{BA}) = - \frac{a}{c - a}\overrightarrow{BA} + \frac{c}{c - a}\overrightarrow{BC} = - \frac{\sqrt{3} - 1}{2}\overrightarrow{BA} + \frac{\sqrt{3} + 1}{2}\overrightarrow{BC}$，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$BD = \frac{\sqrt{2}}{2}$。