《2025年学霸高考黑题数学人教版》

典型例题2. (2023·全国甲理)在△ABC中,∠BAC=60°,AB=2,BC=√6,∠BAC的平分线交BC于D,则AD=2.

答案： 典型例题2.2 解析：记$AB = c$,$AC = b$,$BC = a$,由余弦定理可得,$2^{2} +b^{2} - 2 × 2 × b × \cos 60^{\circ} = 6$.因为$b ＞ 0$,解得$b = 1 + \sqrt{3}$,由$S_{\triangle ABC} = S_{\triangle ABD} +S_{\triangle ACD}$,可得$\frac{1}{2} × 2 × b × \sin 60^{\circ} = \frac{1}{2} × 2 × AD × \sin 30^{\circ} + \frac{1}{2} × AD × b × \sin 30^{\circ}$,
解得$AD = \frac{\sqrt{3}b}{1 + \frac{\sqrt{3}}{2}} = \frac{2\sqrt{3}(1 + \sqrt{3})}{3 + \sqrt{3}} = 2$.

变式训练3. (2025·江苏苏州期末)记△ABC的内角A,B,C的对边分别为a,b,c,已知b+c=a(cos C+√3 sin C),D为边BC上一点,BD=2DC,AD=3.
(1)求A;
(2)若AD平分∠BAC,求a.

答案： 变式训练3. 解：
(1)因为$b + c = a (\cos C + \sqrt{3} \sin C)$,由正弦定理得
$\sin B + \sin C = \sin A (\cos C + \sqrt{3} \sin C)$,则$\cos A \sin C + \sin C = \sqrt{3} \sin C \sin A$.因为$C \in (0,\pi)$,所以
$\sin C \neq 0$,所以$\sqrt{3} \sin A - \cos A = 1$,即$\sin (A - \frac{\pi}{6}) = \frac{1}{2}$.又因为$A \in(0,\pi)$,所以$A = \frac{\pi}{3}$
(2)因为$AD$平分$\angle BAC$,所以$\frac{c}{b} = \frac{BD}{DC} = 2$,即$c = 2b$,由面积相等得$\frac{1}{2} ×3 · 2b · \sin \frac{\pi}{6} + \frac{1}{2} × 3 · b · \sin \frac{\pi}{6} = \frac{1}{2} · b · 2b · \sin \frac{\pi}{3}$,解得$b = \frac{3\sqrt{3}}{2}$,
所以$c = 3\sqrt{3}$.由余弦定理得$a^{2} = (\frac{3\sqrt{3}}{2})^{2} + (3\sqrt{3})^{2} - 2 × \frac{3\sqrt{3}}{2} × 3\sqrt{3} ·\cos \frac{\pi}{3} = \frac{81}{4}$,所以$a = \frac{9}{2}$.

变式训练4. (2025·湖北宜昌模拟)如图所示,在△ABC中,sin C=3 sin B,AD平分∠BAC,且AD=kAC.
(1)若DC=2,求BC的长度;
(2)求k的取值范围;
(3)若S△ABC=3/2,求k为何值时,BC最短.

答案： 变式训练4. 解：
(1)因为$\sin C = 3 \sin B$,由正弦定理得$c = 3b$,在$\triangle ABD$
中,由正弦定理得$\frac{AB}{\sin \angle ADB} = \frac{BD}{\sin \angle BAD}$.在$\triangle ACD$中,由正弦定理得$\frac{AC}{\sin \angle ADC} = \frac{DC}{\sin \angle CAD}$.因为$AD$平分$\angle BAC$,所以$\angle BAD = \angle CAD$.因为
$\angle ADB + \angle ADC = \pi$,所以$\sin \angle ADB = \sin \angle ADC$,所以$\frac{AB}{AC} = \frac{BD}{DC}$.因为$c =3b$,$DC = 2$,所以$\frac{BD}{2} = 3$,得$BD = 6$,所以$BC = 8$.
(2)因为$S_{\triangle ABC} = S_{\triangle ABD} + S_{\triangle ADC}$,设$\angle BAD = \angle CAD = \theta$,所以$\frac{1}{2} AB ·AC \sin 2\theta = \frac{1}{2} AB · AD \sin \theta + \frac{1}{2} AC · AD \sin \theta$.因为$c = 3b$,$AD = kAC$,所以
$3AC · AC · 2 \sin \theta \cos \theta = 3AC · kAC \sin \theta + AC · kAC \sin \theta$.因为$\sin \theta \neq 0$,所
以$6 \cos \theta = 3k + k$,所以$k = \frac{3}{2} \cos \theta$.因为$\theta \in (0,\frac{\pi}{2})$,所以$\cos \theta \in (0,1)$,
所以$k \in (0,\frac{3}{2})$.
(3)由余弦定理得$BC^{2} = c^{2} + b^{2} - 2c · b \cos \angle BAC = 2b^{2}(5 - 3 \cos \angle BAC)$.因
为$S_{\triangle ABC} = \frac{3}{2} \sqrt{3}$,所以$\frac{1}{2} bc \sin 2\theta = \frac{3}{2} \sqrt{3}$.因为$c = 3b$,所以$b^{2} = \frac{1}{\sin \angle BAC}$,所
以$BC^{2} = \frac{2}{\sin \angle BAC} (5 - 3 \cos \angle BAC) = 2 · \frac{5 - 3 \cos \angle BAC}{\sin \angle BAC} = 2 ·\frac{5 - 3 \cos \angle BAC - 1)}{\sin \angle BAC} = \frac{8 - 6 \cos^{2} \theta}{\sin \theta \cos \theta} = \frac{8 \sin^{2} \theta + 2 \cos^{2} \theta}{\sin \theta \cos \theta} = \frac{8 \tan^{2} \theta + 2}{\tan \theta} = 8 \tan \theta + \frac{2}{\tan \theta} \geq 2 \sqrt{8 \tan \theta × \frac{2}{\tan \theta}} = 8$,当且仅当$8 \tan \theta = \frac{2}{\tan \theta}$,即
$\tan \theta = \frac{1}{2}$时等号成立,故此时$\cos \theta = \frac{2\sqrt{5}}{5}$,即$k = \frac{3\sqrt{5}}{5}$.