《2026年南方凤凰台5A新考案数学二轮提高版》

例2 （2025·岳阳质检）已知$a$，$b$，$c$分别为$\triangle ABC$的内角$A$，$B$，$C$的对边，且$\sqrt{3}a\sin C + a\cos C - b = c$，$D$为$BC$边的中点，若$AD = \sqrt{3}$，且$2b^{2} - a^{2} = 4$.
（1）求角$A$的大小；
（2）求$\triangle ABC$的面积.

答案： 例2　[解答]
(1)由$\sqrt{3}a\sin C$ + a cos C - b = c,利用正弦定理可得$\sqrt{3}\sin A\sin C$ + sin A cos C - sin B = sin C,又sin B = sin(A + C) = sin A cos C + cos A sin C,所以$\sqrt{3}\sin A\sin C$ + sin A cos C - sin A cos C - cos A sin C = sin C,即$\sqrt{3}\sin A\sin C$ - cos A sin C = sin C,因为C∈(0,π),所以sin C ＞ 0,故$\sqrt{3}\sin A$ - cos A = 1,由辅助角公式得2sin(A - $\frac{\pi}{6}$) = 1,又A∈(0,π),所以 - $\frac{\pi}{6}$ ＜ A - $\frac{\pi}{6}$ ＜ $\frac{5\pi}{6}$,所以A - $\frac{\pi}{6}$ = $\frac{\pi}{6}$,即A = $\frac{\pi}{3}$.
(2)因为∠ADB + ∠ADC = π,所以cos∠ADB + cos∠ADC = 0,由余弦定理得$\frac{AD^{2}+BD^{2}-AB^{2}}{2AD· BD}$ + $\frac{AD^{2}+DC^{2}-AC^{2}}{2AD· DC}$ = 0,由D为BC的中点,即BD = DC,化简得AC² + AB² = 2(AD² + BD²),又AD = $\sqrt{3}$,所以b² + c² = 2(3 + $\frac{1}{4}a^{2}$),将2b² - a² = 4代入解得c = 2.又cos A = $\frac{1}{2}$ = $\frac{b^{2}+c^{2}-a^{2}}{2bc}$,所以b² + 4 - a² = 2b,将a² = 2b² - 4代入上式得b² + 2b - 8 = 0,解得b = 2(负值舍去),则△ABC的面积为$\frac{1}{2}bc\sin A$ = $\frac{1}{2}$×2×2×$\frac{\sqrt{3}}{2}$ = $\sqrt{3}$.

变式2-1 (2025·聊城一模)在$\triangle ABC$中，已知$AB = 5$，$AC = 4$，$\cos(C - B) = \frac{7}{8}$，则$\triangle ABC$的面积为$\frac{15\sqrt{15}}{8}$.

答案：
变式2 - 1　$\frac{15\sqrt{15}}{8}$　[解析]因为AB ＞ AC,故∠C ＞ ∠B,如图,过点C作射线CD交线段AB于点D,使∠BCD = ∠B,则CD = BD,则cos(C - B) = cos∠ACD = $\frac{7}{8}$.在△ACD中,利用余弦定理得cos∠ACD = $\frac{16 + CD^{2}-(5 - CD)^{2}}{2×4×CD}$ = $\frac{7}{8}$,解得CD = 3,则AD = 2.在△ACD中,利用余弦定理得cos∠CAD = $\frac{16 + 4 - 9}{2×4×2}$ = $\frac{11}{16}$,则sin∠CAD = $\sqrt{1 - (\frac{11}{16})^{2}}$ = $\frac{3\sqrt{15}}{16}$,则S△ABC = $\frac{1}{2}$×4×5×$\frac{3\sqrt{15}}{16}$ = $\frac{15\sqrt{15}}{8}$.
　　 　　　

变式2-2 (2025·南昌一模)三角形是常见的几何图形，除了我们已经学习的性质外，三角形还有很多性质，如
性质1：$\triangle ABC$的面积$S = \frac{1}{2}AB · AC\sin A = \frac{1}{2}\overrightarrow{AB} · \overrightarrow{AC}\tan A$；
性质2：对于$\triangle ABC$内任意一点$P$，有$\overrightarrow{AB} · \overrightarrow{AP} + \overrightarrow{BC} · \overrightarrow{BP} + \overrightarrow{CA} · \overrightarrow{CP} = \overrightarrow{AB} · \overrightarrow{AC} + \overrightarrow{BC} · \overrightarrow{BA} + \overrightarrow{CA} · \overrightarrow{CB}$；
性质3：$\triangle ABC$内存在唯一一点$P$，使得$\angle PAB = \angle PBC = \angle PCA = \alpha$，这个点$P$称为$\triangle ABC$的“勃罗卡点”，角$\alpha$称为$\triangle ABC$的“勃罗卡角”.
若$\triangle ABC$的三边长分别为$1$，$1$，$\sqrt{3}$，根据以上性质，$\triangle ABC$的“勃罗卡角”的正切值为$\frac{\sqrt{3}}{5}$.

答案： 变式2 - 2　$\frac{\sqrt{3}}{5}$　[解析]因为△ABC的三边长分别为1,1,$\sqrt{3}$,不妨设AB = 1,AC = 1,BC = $\sqrt{3}$,如图,由余弦定理得cos A = $\frac{AB^{2}+AC^{2}-BC^{2}}{2×AB×AC}$ = $\frac{1^{2}+1^{2}-(\sqrt{3})^{2}}{2×1×1}$ = - $\frac{1}{2}$,则A = 120°,故B = 30°,C = 30°.在△ABP中,∠APB = 180° - α - (30° - α) = 150°,由正弦定理得$\frac{BP}{\sin\alpha}$ = $\frac{AB}{\sin\angle APB}$,即$\frac{BP}{\sin\alpha}$ = $\frac{1}{\sin150^{\circ}}$ = 2,则BP = 2sinα.在△PBC中,∠BPC = 180° - α - (30° - α) = 150°,由正弦定理得$\frac{BP}{\sin(30^{\circ}-\alpha)}$ = $\frac{BC}{\sin\angle BPC}$,即$\frac{BP}{\sin(30^{\circ}-\alpha)}$ = $\frac{\sqrt{3}}{1}$ = 2$\sqrt{3}$,则2sinα = 2$\sqrt{3}$sin(30° - α),所以sinα = $\sqrt{3}$($\frac{1}{2}\cos\alpha$ - $\frac{\sqrt{3}}{2}\sin\alpha$),则$\frac{5}{2}\sin\alpha$ = $\frac{\sqrt{3}}{2}\cos\alpha$,所以tanα = $\frac{\sqrt{3}}{5}$.