《2025年小题狂做高中数学必修第二册人教版巅峰版》

7. [2024陕西西安期末]咸阳市的标志性建筑清渭楼古朴典雅,蔚为壮观.为了测量清渭楼最高点M与其附近一观测点N之间的距离,取水平方向距离1 200米的P,Q两点,测得∠QPM=π/3,∠QPN=π/4,∠PQM=π/6,∠PQN=5π/12,其中点M,N,P,Q在同一铅垂面内,则M,N两点之间的距离为$200\sqrt {15}$米.

答案： 7. $200\sqrt {15}$由题意知$\angle QPM = \frac {\pi}{3}$,$\angle PQM = \frac {\pi}{6}$,所以$\angle QMP = \frac {\pi}{2}$,所以在$\triangle QMP$中,$MP = \frac {1}{2}QP = 600$米,$QM = \frac {\sqrt {3}}{2}PQ = 600\sqrt {3}$米.又$\angle QPN = \frac {\pi}{4}$,$\angle PQN = \frac {5\pi}{12}$,所以$\angle QNP = \frac {\pi}{3}$,在$\triangle QNP$中,由正弦定理,得$\frac {QN}{\sin \frac {\pi}{4}} = \frac {PQ}{\sin \frac {\pi}{3}}$,所以$QN = \frac {1200}{\frac {\sqrt {3}}{2}} × \frac {\sqrt {2}}{2} = 400\sqrt {6}$(米).在$\triangle MNQ$中,$\angle MQN = \frac {5\pi}{12} - \frac {\pi}{6} = \frac {\pi}{4}$,由余弦定理得$MN = \sqrt {QN^{2} + QM^{2} - 2QN · QM · \cos \angle MQN} = 200\sqrt {15}$(米).

8. [2022山东德州期末]从①√3 b sin A/(1+cos B)=a;②a sin B-√3 b cos B·cos C=√3 c cos² B;③(sin A-sin C)²=sin² B-sin A sin C这三个条件中任选一个,补充在下面问题中,并加以解答.
在△ABC中,a,b,c分别是角A,B,C的对边,若.
(1) 求角B的大小;
(2) 若△ABC为锐角三角形,c=1,求a的取值范围.

答案： 8. 解:
(1)若选①,由正弦定理得$\frac {\sqrt {3} \sin B \sin A}{1 + \cos B} = \sin A$,即$\sqrt {3} \sin B \sin A = \sin A(1 + \cos B)$,因为$0 ＜ A ＜ \pi$,所以$\sin A \neq 0$,所以$\sqrt {3} \sin B = 1 + \cos B$,所以$\sin (B - \frac {\pi}{6}) = \frac {1}{2}$,又因为$- \frac {\pi}{6} ＜ B - \frac {\pi}{6} ＜ \frac {5\pi}{6}$,所以$B = \frac {\pi}{3}$.
若选②,因为$a \sin B - \sqrt {3}b \cos B \cos C = \sqrt {3}c · \cos ^{2}B$,由正弦定理得$\sin A \sin B = \sqrt {3} \sin B · \cos B \cos C + \sqrt {3} \sin C · \cos ^{2}B$,即$\sin A \sin B = \sqrt {3} \cos B(\sin B \cos C + \sin C \cos B) = \sqrt {3} \cos B \sin (B + C)$,所以$\sin A \sin B = \sqrt {3} \cos B \sin A$,由$A \in (0, \pi)$,得$\sin A \neq 0$,所以$\sin B = \sqrt {3} \cos B$,即$\tan B = \sqrt {3}$,因为$B \in (0, \pi)$,所以$B = \frac {\pi}{3}$.
若选③,由$(\sin A - \sin C)^{2} = \sin ^{2}B - \sin A \sin C$,化简得$\sin ^{2}A + \sin ^{2}C - \sin ^{2}B = \sin A \sin C$.由正弦定理得$a^{2} + c^{2} - b^{2} = ac$,即$\frac {a^{2} + c^{2} - b^{2}}{2ac} = \frac {1}{2}$,所以$\cos B = \frac {1}{2}$.因为$B \in (0, \pi)$,所以$B = \frac {\pi}{3}$.
(2)在$\triangle ABC$中,由正弦定理$\frac {a}{\sin A} = \frac {c}{\sin C}$,得$a = \frac {c \sin A}{\sin C}$,由
(1)知$B = \frac {\pi}{3}$,又$c = 1$,代入上式得$a = \frac {\sin (\frac {2\pi}{3} - C)}{\sin C} = \frac {\frac {1}{2} \sin C + \frac {\sqrt {3}}{2} \cos C}{\sin C} = \frac {1}{2} + \frac {\sqrt {3}}{2 \tan C}$.因为$\begin{cases} 0 ＜ C ＜ \frac {\pi}{2}, \\ 0 ＜ \frac {2\pi}{3} - C ＜ \frac {\pi}{2}, \end{cases} \triangle ABC$为锐角三角形,所以$\begin{cases} 0 ＜ C ＜ \frac {\pi}{2}, \\ \frac {\pi}{6} ＜ C ＜ \frac {\pi}{2}, \end{cases}$所以$\tan C ＞ \frac {\sqrt {3}}{3}$,所以$a = \frac {1}{2} + \frac {\sqrt {3}}{2 \tan C} \in (\frac {1}{2}, 2)$.

9. [2025浙江温州期末]已知△ABC的内角A,B,C的对边为a,b,c,M是△ABC的内心,若a=2,√3 b cos A=a sin B.
(1) 求角A的大小;
(2) 延长AM交BC于点D,若AD=2√3/3,求△ABC的周长;
(3) 求AM的取值范围.

答案： 9. 解:
(1)因为$\sqrt {3}b \cos A = a \sin B$,所以由正弦定理可得$\sqrt {3} \sin B \cos A = \sin A \sin B$,因为$B \in (0, \pi)$,所以$\sin B ＞ 0$,所以$\tan A = \sqrt {3}$,因为$A \in (0, \pi)$,所以$A = \frac {\pi}{3}$.
(2)因为$M$是$\triangle ABC$的内心,延长$AM$交$BC$于点$D$,所以$AD$为$\angle BAC$的平分线.因为$AD = \frac {2\sqrt {3}}{3}$,所以$S_{\triangle ABC} = \frac {1}{2}bc \sin A = \frac {1}{2}(b + c) · AD · \sin \frac {A}{2}$,所以$bc · \frac {\sqrt {3}}{2} = (b + c) · \frac {2\sqrt {3}}{3} · \frac {1}{2}$,可得$3bc = 2(b + c)$.由余弦定理可得$a^{2} = b^{2} + c^{2} - 2bc \cos A = (b + c)^{2} - 3bc = (b + c)^{2} - 2(b + c)$,即$(b + c)^{2} - 2(b + c) - 4 = 0$,解得$b + c = \frac {2 + \sqrt {4 + 4 × 4}}{2} = 1 + \sqrt {5}$(负值已舍),所以$\triangle ABC$的周长为$a + b + c = 3 + \sqrt {5}$.
(3)记$\triangle ABC$内切圆的半径为$r$,由等面积思想有$\frac {1}{2}(a + b + c) · r = \frac {1}{2}bc \sin 60^{\circ}$,可得$r = \frac {\sqrt {3}}{2} · \frac {bc}{b + c + 2}$,则$AM = 2r = \sqrt {3} · \frac {bc}{b + c + 2}$,由余弦定理得$b^{2} + c^{2} - bc = 4$,可知$(b + c)^{2} - 4 = 3bc$,则$AM = \frac {\sqrt {3}}{3} · \frac {(b + c)^{2} - 4}{b + c + 2} = \frac {\sqrt {3}}{3} · (b + c - 2)$,又因为$(b + c)^{2} - 4 = 3bc \leq 3(\frac {b + c}{2})^{2}$,解得$2 ＜ b + c \leq 4$,所以$AM = \frac {\sqrt {3}}{3} · (b + c - 2) \in (0, \frac {2\sqrt {3}}{3}]$.