《2026年南方凤凰台5A新考案数学二轮提高版》

例1 如图，已知$P$是$\triangle ABC$内一点，$PB = PC$，$\angle BAC = \frac{\pi}{4}$，$\angle BPC = \frac{3\pi}{4}$，$\angle ABP = \theta$.
（1）若$\theta = \frac{\pi}{24}$，$BC = \sqrt{2}$，求$AC$的长；
（2）若$\theta = \frac{\pi}{3}$，求$\tan \angle BAP$的值.（1）1；（2）$3 - \sqrt{6}$

答案：
例1　[解答]
(1)如图
(1),在△BPC中,∠BPC = $\frac{3\pi}{4}$,PB = PC,所以∠PBC = $\frac{\pi}{8}$,所以∠ABC = ∠PBC + θ = $\frac{\pi}{8}$ + $\frac{\pi}{24}$ = $\frac{\pi}{6}$.在△ABC中,由正弦定理得$\frac{AC}{\sin\angle ABC}$ = $\frac{BC}{\sin\angle BAC}$,即$\frac{AC}{\frac{1}{2}}$ = $\frac{\sqrt{2}}{\frac{\sqrt{2}}{2}}$,解得AC = 1.
　　
(2)如图
(2),当θ = $\frac{\pi}{3}$时,∠ACP = π - ∠BAC - ∠ABP - 2∠PBC = $\frac{\pi}{6}$.设∠BAP = α,则∠PAC = $\frac{\pi}{4}$ - α.在△ABP中,由正弦定理得$\frac{AP}{PB}$ = $\frac{\sin\frac{\pi}{3}}{\sin\alpha}$.在△APC中,由正弦定理得$\frac{AP}{PC}$ = $\frac{\sin(\frac{\pi}{4}-\alpha)}{\sin\frac{\pi}{6}}$.因为PB = PC,所以$\frac{\sin\frac{\pi}{3}}{\sin\alpha}$ = $\frac{\sin(\frac{\pi}{4}-\alpha)}{\sin\frac{\pi}{6}}$,即$\frac{\frac{\sqrt{3}}{2}}{\sin\alpha}$ = $\frac{\frac{\sqrt{2}}{2}(\cos\alpha - \sin\alpha)}{\frac{1}{2}}$,整理得$\frac{\sqrt{3}}{\sin\alpha}$ = $\frac{\sqrt{2}}{\cos\alpha - \sin\alpha}$,即$\frac{\sqrt{3}}{\tan\alpha}$ = $\frac{\sqrt{2}}{1 - \tan\alpha}$,解得tanα = 3 - $\sqrt{6}$,即tan∠BAP = 3 - $\sqrt{6}$.

变式1-1 在平面四边形$ABCD$中，$AD = 1$，$AB = CD = 2$，$BC = 3$，且四边形$ABCD$的面积为$\sqrt{3}$，则$\cos(\angle ADC + \angle ABC) = $$\frac{1}{2}$.

答案：
变式1 - 1　$\frac{1}{2}$　[解析]如图,根据题意,连接AC,在△ADC中,AD = 1,DC = 2,由余弦定理得AC² = AD² + DC² - 2AD·DCcosD = 1 + 4 - 2×1×2cosD = 5 - 4cosD.在△ABC中,AB = 2,BC = 3,由余弦定理得AC² = AB² + BC² - 2AB·BC·cosB = 4 + 9 - 2×2×3cosB = 13 - 12cosB,所以5 - 4cosD = 13 - 12cosB,化简得cosD - 3cosB = - 2①.S四边形ABCD = S△ADC + S△ABC = $\frac{1}{2}$×1×2sinD + $\frac{1}{2}$×2×3sinB = $\sqrt{3}$,化简得sinD + 3sinB = $\sqrt{3}$②.由①² + ②²,可得(cosD - 3cosB)² + (sinD + 3sinB)² = 7,整理得(cos²D + sin²D) + 9(cos²B + sin²B) - 6(cosBcosD - sinBsinD) = 7,即10 - 6cos(B + D) = 7,解得cos(B + D) = $\frac{1}{2}$,即cos(∠ADC + ∠ABC) = $\frac{1}{2}$.
　　　　　　　　

变式1-2 （2025·襄阳模拟）在$\triangle ABC$中，内角$A$，$B$，$C$的对边分别为$a$，$b$，$c$，$BC = 1$且$b\cos C + \sqrt{3}c\sin B = 1 + 2c$.
（1）求角$B$的大小；
（2）如图，$D$为$\triangle ABC$外一点，$\angle DCB = \angle B$，$CD = \sqrt{3}$，$AC = AD$，求$\sin \angle BCA$的值.
（1）$120^{\circ}$；（2）$\frac{\sqrt{2}}{2}$

答案： 变式1 - 2　[解答]
(1)因为bcosC + $\sqrt{3}c\sin B$ = 1 + 2c = a + 2c,所以在△ABC中,由正弦定理得sinBcosC + $\sqrt{3}\sin C\sin B$ = sinA + 2sinC,由三角形内角和为180°可得sinA = sin(B + C) = sin B·cosC + cosBsinC,所以sinBcosC + $\sqrt{3}\sin C\sin B$ = sin(B + C) + 2sinC = sinB·cosC + cosBsinC + 2sinC,即$\sqrt{3}\sin C\sin B$ - cosBsinC = 2sinC,因为0° ＜ C ＜ 180°,所以sinC ≠ 0,所以$\sqrt{3}\sin B$ - cosB = 2,即$\frac{\sqrt{3}}{2}\sin B$ - $\frac{1}{2}\cos B$ = 1,所以sin(B - 30°) = 1,又因为0° ＜ B ＜ 180°,所以B - 30° = 90°,即B = 120°.
(2)因为AC = AD,令∠DCA = ∠CDA = α,则∠CAD = 180° - 2α,在△ACD中,由正弦定理得$\frac{AC}{\sin D}$ = $\frac{CD}{\sin\angle CAD}$,又CD = $\sqrt{3}$,所以AC = $\frac{\sqrt{3}\sin\alpha}{\sin(180^{\circ}-2\alpha)}$ = $\frac{\sqrt{3}}{2\cos\alpha}$.在△ABC中,由正弦定理得$\frac{AC}{\sin B}$ = $\frac{BC}{\sin\angle BAC}$,又∠BAC = 180° - 120° - (120° - α) = α - 60°,BC = 1,所以AC = $\frac{\sin120^{\circ}}{\sin(\alpha-60^{\circ})}$,所以sin(α - 60°) = cosα = sin(90° - α),解得α = 75°,所以sin∠BCA = sin(120° - 75°) = $\frac{\sqrt{2}}{2}$.