《2025年5年高考3年模拟高中数学全一册人教B版》

2.(2024湖北鄂东南省级高中联考,15)记△ABC的角A,B,C的对边分别为a,b,c,已知$\frac{\sin A - \sin B}{b + c}$ = $\frac{\sin C}{a + b}$.
(1)求角A;
(2)若点D是BC边上一点,且AB⊥AD,CD = 2BD,求sin∠ADB的值.

答案： 解析：
(1)由$\frac{\sin A-\sin B}{b + c}=\frac{\sin C}{a + b}$及正弦定理得$\frac{a - b}{b + c}=\frac{c}{a + b}$，即$a^{2}=b^{2}+c^{2}+bc$。又$a^{2}=b^{2}+c^{2}-2bc\cos A$，$\therefore\cos A=-\frac{1}{2}$，$\therefore A=\frac{2\pi}{3}$。
(2)$\angle DAC=\angle BAC-\angle BAD=\frac{\pi}{6}$，记$\angle ADB=\alpha$，$\alpha\in(0,\frac{\pi}{2})$，则$C=\alpha-\angle DAC=\alpha-\frac{\pi}{6}$。在$Rt\triangle ABD$中，$AD = BD\cos\alpha$。①在$\triangle ADC$中，由正弦定理得$\frac{AD}{\sin(\alpha-\frac{\pi}{6})}=\frac{CD}{\sin\frac{\pi}{6}}$。②由①②及$CD = 2BD$得$\frac{\cos\alpha}{\sin(\alpha-\frac{\pi}{6})}=\frac{2}{\sin\frac{\pi}{6}}$，即$\frac{1}{\frac{\sqrt{3}}{2}\tan\alpha-\frac{1}{2}} = 4$，解得$\tan\alpha=\frac{\sqrt{3}}{2}$。由$\tan\alpha=\frac{\sqrt{3}}{2}$，$\sin^{2}\alpha+\cos^{2}\alpha = 1$，$\alpha\in(0,\frac{\pi}{2})$，解得$\sin\alpha=\frac{\sqrt{21}}{7}$。故$\sin\angle ADB=\frac{\sqrt{21}}{7}$。

3.(2024山东济南二模,15)如图,在平面四边形ABCD中,BC⊥CD,AB = BC = $\sqrt{2}$,∠ABC = θ,120° ≤ θ ＜ 180°.
(1)若θ = 120°,AD = 3,求∠ADC的大小;
(2)若CD = $\sqrt{6}$,求四边形ABCD面积的最大值.

答案： 解析：
(1)因为$\angle ABC = 120^{\circ}$，$AB = BC=\sqrt{2}$，所以$\angle ACB=\angle BAC = 30^{\circ}$，$AC = 2AB\cos30^{\circ}=\sqrt{6}$，又$BC\perp CD$，所以$\angle ACD = 90^{\circ}-\angle ACB = 60^{\circ}$，在$\triangle ACD$中，由正弦定理得$\frac{AD}{\sin\angle ACD}=\frac{AC}{\sin\angle ADC}$，所以$\sin\angle ADC=\frac{AC\sin\angle ACD}{AD}=\frac{\sqrt{6}\sin60^{\circ}}{3}=\frac{\sqrt{2}}{2}$，又$AC＜AD$，所以$\angle ADC = 45^{\circ}$。
(2)在$\triangle ABC$中，$AB = BC=\sqrt{2}$，$\angle ABC=\theta$，$S_{\triangle ABC}=\frac{1}{2}AB\cdot BC\cdot\sin\angle ABC=\frac{1}{2}\times\sqrt{2}\times\sqrt{2}\times\sin\theta=\sin\theta$，$\frac{AC}{2}=\sin\frac{\theta}{2}\Rightarrow AC = 2\sqrt{2}\sin\frac{\theta}{2}$，在$\triangle ACD$中，$\angle ACD = 90^{\circ}-\angle ACB = 90^{\circ}-\frac{180^{\circ}-\theta}{2}=\frac{\theta}{2}$，又$CD=\sqrt{6}$，所以$S_{\triangle ACD}=\frac{1}{2}AC\cdot CD\cdot\sin\angle ACD=\frac{1}{2}\cdot2\sqrt{2}\sin\frac{\theta}{2}\cdot\sqrt{6}\cdot\sin\frac{\theta}{2}=\sqrt{3}-\sqrt{3}\cos\theta$，所以四边形$ABCD$的面积$S = S_{\triangle ABC}+S_{\triangle ACD}=\sin\theta+\sqrt{3}-\sqrt{3}\cos\theta=\sqrt{3}+2\sin(\theta - 60^{\circ})$，因为$120^{\circ}\leqslant\theta＜180^{\circ}$，所以$60^{\circ}\leqslant\theta - 60^{\circ}＜120^{\circ}$，所以当$\theta - 60^{\circ}=90^{\circ}$，即$\theta = 150^{\circ}$时，$S_{max}=\sqrt{3}+2$，故四边形$ABCD$面积的最大值为$\sqrt{3}+2$。

4.(2024福建九地市质量检测,15)△ABC中,角A,B,C的对边分别是a,b,c,且asinC = csinB,C = $\frac{2\pi}{3}$.
(1)求B的大小;
(2)若△ABC的面积为$\frac{3\sqrt{3}}{4}$,求BC边上中线的长.

答案： 解析：
(1)$\because a\sin C = c\sin B$，$\therefore$由正弦定理，得$\sin A\sin C=\sin C\sin B$，$\because 0＜C＜\pi$，$\therefore\sin C＞0$，$\therefore\sin A=\sin B$，$\because 0＜A＜\pi$，$0＜B＜\pi$，$\therefore A = B$或$A + B=\pi$（舍去），$\because A + B + C=\pi$，且$C=\frac{2\pi}{3}$，$\therefore B=\frac{\pi}{6}$。
(2)依题意得$\frac{3\sqrt{3}}{4}=\frac{1}{2}ab\sin C$，$\because A = B$，$\therefore a = b$，$\therefore\frac{3\sqrt{3}}{4}=\frac{1}{2}a^{2}\sin\frac{2\pi}{3}=\frac{\sqrt{3}a^{2}}{4}$，解得$a=\sqrt{3}$，设边$BC$的中点为$D$，则$CD=\frac{\sqrt{3}}{2}$，在$\triangle ACD$中，由余弦定理得$AD^{2}=AC^{2}+CD^{2}-2AC\cdot CD\cos C=3+\frac{3}{4}-2\times\sqrt{3}\times\frac{\sqrt{3}}{2}\times\cos\frac{2\pi}{3}=\frac{21}{4}$，$\therefore BC$边上中线的长为$\frac{\sqrt{21}}{2}$。

5.(2024江西重点中学协作体联考,16)在△ABC中,内角A,B,C所对的边分别为a,b,c,其外接圆的半径为2$\sqrt{3}$,且bcosC = a + $\frac{\sqrt{3}}{3}$csinB.
(1)求角B;
(2)若∠ABC的平分线交AC于点D,BD = $\sqrt{3}$,点E在线段AC上,EC = 2EA,求△BDE的面积.

答案： 解析：
(1)由正弦定理可得$\sin B\cos C=\sin A+\frac{\sqrt{3}}{3}\sin C\sin B$。又$\sin A=\sin(B + C)=\sin B\cos C+\cos B\sin C$，则$\sin C\cos B+\frac{\sqrt{3}}{3}\sin C\sin B = 0$。$\because C\in(0,\pi)$，$\therefore\cos B+\frac{\sqrt{3}}{3}\sin B = 0$，即$\tan B=-\sqrt{3}$。又$B\in(0,\pi)$，$\therefore B=\frac{2\pi}{3}$。
(2)由
(1)可知$B=\frac{2\pi}{3}$，$\because\triangle ABC$的外接圆的半径为$2\sqrt{3}$，$\therefore$由正弦定理得$\frac{b}{\sin B}=4\sqrt{3}$，即$b = 6$，$\because BD$平分$\angle ABC$，$\therefore\angle CBD=\angle ABD=\frac{1}{2}\angle ABC=\frac{\pi}{3}$。由$S_{\triangle ABC}=S_{\triangle BCD}+S_{\triangle ABD}$，可得$\frac{1}{2}ac\sin\frac{2\pi}{3}=\frac{1}{2}a\cdot\sqrt{3}\sin\frac{\pi}{3}+\frac{1}{2}c\cdot\sqrt{3}\sin\frac{\pi}{3}$，即$ac=\sqrt{3}(a + c)$①，由余弦定理得$b^{2}=a^{2}+c^{2}-2ac\cos\frac{2\pi}{3}$，即$(a + c)^{2}-ac = 36$②，由①②可得$a = c = 2\sqrt{3}$。则$BD\perp AC$，$\therefore AC = 2CD = 6$，又$\because EC = 2AE$，则$DE = 1$，故$S_{\triangle BDE}=\frac{1}{2}\times1\times\sqrt{3}=\frac{\sqrt{3}}{2}$。