《2026年南方凤凰台5A新考案数学二轮提高版》

1. 在$\triangle ABC$中，$\angle B = 45^{\circ}$，$D$是$BC$边上一点，$AD = 5$，$AC = 7$，$DC = 3$，则$AB$的长为(C)

A.$5\sqrt{2}$
B.$3\sqrt{6}$
C.$\frac{5\sqrt{6}}{2}$
D.$4\sqrt{3}$

答案：
1.C　[解析]如图所示,在△ADC中,AD = 5,AC = 7,DC = 3,由余弦定理得cos∠ADC = $\frac{AD^{2}+DC^{2}-AC^{2}}{2AD· DC}$ = $\frac{25 + 9 - 49}{2×5×3}$ = -$\frac{1}{2}$,所以∠ADC = 120°,∠ADB = 60°.在△ABD中,AD = 5,∠B = 45°,∠ADB = 60°,由正弦定理得$\frac{AB}{\sin\angle ADB}$ = $\frac{AD}{\sin B}$,所以AB = $\frac{5×\sin60^{\circ}}{\sin45^{\circ}}$ = $\frac{5\sqrt{6}}{2}$.
　　　　　　　　

2. 如图，在四边形$ABCD$中，$AC \perp BC$，$AB = 4$，$BC = CD$，$\angle ACD = 60^{\circ}$，则$AD$长的最小值为(C)

A.$2$
B.$\sqrt{6}$
C.$2\sqrt{2}$
D.$2\sqrt{3}$

答案： 2.C　[解析]设∠BAC = α,则α∈(0,$\frac{\pi}{2}$),AC = 4cosα,CD = BC = 4sinα,则在△ACD中利用余弦定理得,AD = $\sqrt{AC^{2}+CD^{2}-2AC· CD·\cos60^{\circ}}$ = $\sqrt{16 - 8\sin2\alpha}$,当sin2α = 1,即α = $\frac{\pi}{4}$时,AD取得最小值2$\sqrt{2}$.

3. 在$\triangle ABC$中，$\angle ACB = \frac{2\pi}{3}$，$D$为$AB$上一点，$CD \perp AC$，$AD = 2DB = 2$，则$\triangle ABC$的面积为(B)

A.$\frac{\sqrt{3}}{2}$
B.$\frac{3\sqrt{3}}{4}$
C.$\frac{3}{4}$
D.$\frac{3}{2}$

答案：
3.B　[解析]根据∠ACB = $\frac{2\pi}{3}$,CD⊥AC,可得∠BCD = $\frac{2\pi}{3}$ - $\frac{\pi}{2}$ = $\frac{\pi}{6}$.在△ABC中,设角A,B,C的对边分别为a,b,c,可得$\frac{AD}{BD}$ = $\frac{S_{\triangle ACD}}{S_{\triangle BCD}}$ = $\frac{\frac{1}{2}b· CD}{\frac{1}{2}a· CD\sin\angle BCD}$ = 2,结合sin∠BCD = $\frac{1}{2}$,化简得a = b.根据余弦定理,可得c² = a² + b² - 2abcos$\frac{2\pi}{3}$ = a² + b² + ab = 3a²,因为AD = 2DB = 2,所以DB = 1,可得c = AD + DB = 3,所以3a² = c² = 9,解得a = $\sqrt{3}$,即a = b = $\sqrt{3}$,所以△ABC的面积S = $\frac{1}{2}ab\sin\angle ACB$ = $\frac{1}{2}$×$\sqrt{3}$×$\sqrt{3}$×sin$\frac{2\pi}{3}$ = $\frac{3\sqrt{3}}{4}$.
　　　　　　　

4. 如图，无人机在离地面高$100\ m$的$A$处，观测到山顶$M$处的仰角为$15^{\circ}$，山脚$C$处的俯角为$45^{\circ}$.已知$\angle MCN = 60^{\circ}$，则山的高度$MN$为(B)

A.$100\sqrt{2}\ m$
B.$150\ m$
C.$150\sqrt{2}\ m$
D.$150\sqrt{3}\ m$

答案： 4.B　[解析]由题知,∠MAD = 15°,∠CAD = 45°,在Rt△ACE中,易知∠CAE = 45°,AE = 100m,所以AC = 100$\sqrt{2}$m.又因为∠CAM = ∠CAD + ∠MAD = 45° + 15° = 60°,∠MCN = 60°,所以∠MCA = 75°,∠CMA = 45°.在△ACM中,由正弦定理得$\frac{AC}{\sin\angle CMA}$ = $\frac{CM}{\sin\angle CAM}$,即$\frac{100\sqrt{2}}{\sin45^{\circ}}$ = $\frac{CM}{\sin60^{\circ}}$,所以CM = 100$\sqrt{3}$m.在Rt△MNC中,MN = CMsin∠MCN = 100$\sqrt{3}$×$\frac{\sqrt{3}}{2}$ = 150(m).