《2025年学霸高考黑题数学人教版》

典型例题1. (2023·新课标全国Ⅱ)记△ABC的内角A,B,C的对边分别为a,b,c,已知△ABC的面积为√3,D为BC中点,且AD=1.
(1)若∠ADC=π/3,求tan B;
(2)若b²+c²=8,求b,c.

答案： 典型例题1. 解：
(1) 在$\triangle ABC$中.因为D为BC中点$,\angle ADC = \frac{\pi}{3},AD = 1,$则$S_{\triangle ADC} = \frac{1}{2}AD · DC \sin \angle ADC = \frac{1}{2} × 1 × \frac{1}{2} a × \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8} a = \frac{1}{2} S_{\triangle ABC} = \frac{\sqrt{3}}{2},$解得a = 4. 在$\triangle ABD$中，$\angle ADB = \frac{2\pi}{3},$由余弦定理得$c^{2} = BD^{2} + AD^{2} - 2BD · AD \cos \angle ADB,$即$c^{2} = 4 + 1 - 2 × 2 × 1 × (-\frac{1}{2}) = 7,$解得c =
一学霸高考·黑
$\sqrt{21}$
14
,所以$\tan B = \frac{\sin B}{\cos B} = \frac{\sqrt{3}}{5}.$
(2)方法一:在$\triangle ABD$与$\triangle ACD$中,由余弦定理得
$c^{2} = \frac{1}{4} a^{2} + 2 = b^{2} + c^{2},$
$b^{2} = \frac{1}{4} a^{2} + 1 - 2 × \frac{1}{2} a × 1 × \cos (\pi - \angle ADC),$
整理得$\frac{1}{2} a^{2} + 2 = b^{2} + c^{2},$而
$b^{2} + c^{2} = 8,$则$a = 2\sqrt{3},$又$S_{\triangle ADC} = \frac{1}{2} × \sqrt{3} × 1 × \sin \angle ADC = \frac{\sqrt{3}}{2},$解得
$\sin \angle ADC = 1,$而$0 ＜ \angle ADC ＜ \pi,$于是$\angle ADC = \frac{\pi}{2},$所以b = c =
$\sqrt{AD^{2} + CD^{2}} = 2.$
方法二:在$\triangle ABC$中,因为D为BC中点,所以$2\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{AC},$又$\overrightarrow{CB} = \overrightarrow{AB} - \overrightarrow{AC},$于是$4\overrightarrow{AD}^{2} + \overrightarrow{CB}^{2} = (\overrightarrow{AB} + \overrightarrow{AC})^{2} + (\overrightarrow{AB} - \overrightarrow{AC})^{2} = 2(b^{2} + c^{2}) = 16,$即
$4 + a^{2} = 16,$解得$a = 2\sqrt{3},$又$S_{\triangle ADC} = \frac{1}{2} × \sqrt{3} × 1 × \sin \angle ADC = \frac{\sqrt{3}}{2},$解得
$\sin \angle ADC = 1,$而$0 ＜ \angle ADC ＜ \pi,$于是$\angle ADC = \frac{\pi}{2},$所以b = c =
$\sqrt{AD^{2} + CD^{2}} = 2.$

变式训练1. (2025·四川成都模拟)已知a,b,c分别为△ABC三个内角A,B,C的对边,(sin B-sin C)²=sin²(B+C)-sin B sin C,且△ABC的面积为2√3.
(1)求A;
(2)若$\overrightarrow{BD}=2\overrightarrow{DC}$,求AD的最小值.

答案： 变式训练1. 解：
(1) 由$(\sin B - \sin C)^{2} = \sin^{2}(B + C) - \sin B \sin C$,可
得$\sin^{2}B - 2\sin B \sin C + \sin^{2}C = \sin^{2}(\pi - A) - \sin B \sin C$,所以$\sin^{2}B -\sin B \sin C + \sin^{2}C = \sin^{2}A$,所以$\sin B \sin C + \sin^{2}C - \sin^{2}A = \sin B \sin C$,由正弦定
理可得$b^{2} + c^{2} - a^{2} = bc$,由余弦定理可得$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc} = \frac{1}{2}$.因为
$A \in (0,\pi)$,所以$A = \frac{\pi}{3}$
(2)因为$\triangle ABC$的面积为$2\sqrt{3}$,所以$\frac{1}{2}bc \sin A = \frac{1}{2} bc \sin \frac{\pi}{3} = 2\sqrt{3}$,所以
$bc = 8$,由$\overrightarrow{BD} = 2\overrightarrow{DC}$,可得$\overrightarrow{BD} = \frac{2}{3} \overrightarrow{BC} = \frac{2}{3} (\overrightarrow{AC} - \overrightarrow{AB})$,所以$\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BD} = \overrightarrow{AB} + \frac{2}{3} (\overrightarrow{AC} - \overrightarrow{AB}) = \frac{1}{3} \overrightarrow{AB} + \frac{4}{9} \overrightarrow{AB} · \overrightarrow{AC} + \frac{4}{9} \overrightarrow{AC}^{2} = \frac{1}{9} c^{2} + \frac{4}{9} cb \cos \frac{\pi}{3} + \frac{4}{9} b^{2} = \frac{1}{9} c^{2} + \frac{2}{9} × 8 + \frac{4}{9} b^{2} = \frac{1}{9} c^{2} + \frac{4}{9} b^{2} + \frac{16}{9} \geq 2 \sqrt{\frac{1}{9} c^{2} × \frac{4}{9} b^{2}} + \frac{16}{9} = \frac{48}{9}$,所以
$|AD| \geq \frac{4\sqrt{3}}{3}$,当且仅当$\frac{1}{9} c^{2} = \frac{4}{9} b^{2}$,即$b = 2$,$c = 4$时取等号,所以$AD$的最小值为
$\frac{4\sqrt{3}}{3}$

变式训练2. (2025·河北石家庄三模)设△ABC的内角A,B,C的对边分别为a,b,c,已知b sin C+√3c cos B=√3a.
(1)求角C的大小;
(2)若AB=2√2,且sin A sin B=3/8,求AB边上中线CT的长.

答案： 变式训练2. 解：
(1) 在$\triangle ABC$中,由$b \sin C + \sqrt{3} c \cos B = \sqrt{3} a$及正弦定理
得$\sin B \sin C + \sqrt{3} \sin C \cos B = \sqrt{3} \sin A = \sqrt{3} \sin (B + C) = \sqrt{3} \sin B \cos C +\sqrt{3} \cos B \sin C$,即$\sin B \sin C = \sqrt{3} \sin B \cos C$.因为$B,C \in (0,\pi)$,则$\sin B ＞0$,即$\sin C = \sqrt{3} \cos C ＞ 0$,可得$\tan C = \sqrt{3}$,故$C = \frac{\pi}{3}$
(2)由正弦定理可得$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = \frac{2\sqrt{2}}{\sin \frac{\pi}{3}} = \frac{4\sqrt{6}}{3}$,所以$ab = \sin A \sin B · \frac{c^{2}}{\sin^{2}C} = 4$,在$\triangle ABC$中,由余弦定理可得$c^{2} = 8 = a^{2} + b^{2} -2ab \cos C = a^{2} + b^{2} - 4$,所以$a^{2} + b^{2} = 12$.因为$CT$为$AB$边上的
中线,所以$\overrightarrow{CT} = \frac{1}{2} (\overrightarrow{CB} + \overrightarrow{CA})$,所以$\overrightarrow{CT}^{2} = \frac{1}{4} (\overrightarrow{CB} + \overrightarrow{CA})^{2} = \frac{1}{4} (\overrightarrow{CB}^{2} +\overrightarrow{CA}^{2} + 2\overrightarrow{CB} · \overrightarrow{CA}) = \frac{1}{4} (a^{2} + b^{2} + 2ab \cos C) = \frac{1}{4} (a^{2} + b^{2} + ab) = \frac{1}{4} × (12 +4) = 4$,故$|CT| = 2$,因此,$AB$边上的中线$CT$的长为$2$.