6.4.3(2) 正弦定理
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必修第二册同步巩固,难度2颗星!
基础知识
正弦定理
(1)内容
在一个三角形中,各边和它所对角的正弦的比相等,即
\(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2 R\) (其中\(R\)是三角形外接圆半径)
证明 设\(∆ABC\)的外接圆为\(⊙O\),
① 当\(∆ABC\)是直角三角形,\(∠C=\dfrac{\pi}{2}\),因为 \(\sin A=\dfrac{B C}{A B}=\dfrac{a}{2 R}\),所以 \(\dfrac{a}{\sin A}=2 R\),
同理 \(\dfrac{b}{\sin B}=2 R\),而 \(\dfrac{c}{\sin C}=\dfrac{2 R}{\sin \dfrac{\pi}{2}}=2 R\),
所以 \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2 R\);
② 当\(∆ABC\)是锐角三角形,过点\(A\)作直径\(AD\),连接\(CD\),
则 \(\dfrac{b}{\sin B}=\dfrac{b}{\sin D}=2 R\),同理 \(\dfrac{a}{\sin A}=2 R\), \(\dfrac{c}{\sin C}=2 R\),
所以\(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2 R\);
③ 当\(∆ABC\)是钝角三角形,令\(∠C>\dfrac{\pi}{2}\),优弧 \(\widehat{A B}\)上取点\(D\),
与\(∆ABC\)是锐角三角形时方法一样可得\(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=2 R\);
而 \(\dfrac{c}{\sin C}=\dfrac{c}{\sin (\pi-D)}=\dfrac{c}{\sin D}=2 R\),所以\(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2 R\);
综上可得任意三角形中\(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2 R\) (其中\(R\)是三角形外接圆半径).
PS 证明正弦定理有很多方法,课本中是向量法;这里使用外圆法,主要是引出比值等于\(2R\),有助于对后面变式的理解.
(2)变形
① \(\dfrac{a+b+c}{\sin A+\sin B+\sin C}=\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\)
② 化边为角
\(a=2 R \sin A\) ,\(b=2R \sin B\) ,\(c=2R \sin C\)
\(a∶ b∶ c=\sin A∶ \sin B∶ \sin C\) ,
\(\dfrac{a}{b}=\dfrac{\sin A}{\sin B}\) , \(\dfrac{b}{c}=\dfrac{\sin B}{\sin C}\) ,\(\dfrac{a}{c}=\dfrac{\sin A}{\sin C}\)
③ 化角为边
\(\sin A=\dfrac{a}{2 R}\),\(\sin B=\dfrac{b}{2 R}\) , \(\sin C=\dfrac{c}{2 R}\)
\(\dfrac{\sin A}{\sin B}=\dfrac{a}{b}\) , \(\dfrac{\sin B}{\sin C}=\dfrac{b}{c}\) , \(\dfrac{\sin A}{\sin C}=\dfrac{a}{c}\)
(3) 正弦定理的“齐次角边互换”
等式\((*)\)中含有三个式子\((a⋅\sin B、c⋅\sin C、b\cdot \sin C)\),每个式子中都有一个\(\sin\)值,并且它们的次数都是\(1\),则可以把\(\sin B\)、\(\sin C\)直接转化为对应的边\(b\)、\(c\)!
同理\(a⋅\sin B+c⋅\sin C=b\cdot \sin C⇒ \sin A⋅\sin B+\sin C⋅\sin C=\sin B⋅\sin C\).
思考以下转化是否正确
(1)\(a\cdot \sin B+c\cdot \sin C=b⇒a\cdot b+c\cdot c=b\) (错),
(2) \(\sin A\cdot \sin B+\sin B\cdot \sin C=\sin ^2A⇒a\cdot b+b\cdot c=a^2\) (对)
利用正弦定理可以解决下列两类三角形的问题
(1) 已知两个角及任意—边,求其他两边和另一角;
【例】 在\(△ABC\),内角\(A\) ,\(B\) ,\(C\)所对的边分别是\(a\),\(b\),\(c\),\(B=30^{\circ}\),\(A=45^{\circ}\),\(b=2\),则边\(a=\) \(\underline{\quad \quad}\) .
答案 \(a=2\sqrt{2}\).
(2) 已知两边和其中—边的对角,求其他两个角及另一边.
【例】 在\(△ABC\),内角\(A\) ,\(B\) ,\(C\)所对的边分别是\(a\),\(b\),\(c\),\(A=60^{\circ}\) ,\(c=\sqrt{2}\),\(a=\sqrt{3}\),则角\(C=\)\(\underline{\quad \quad}\) .
答案 \(C=\dfrac{\pi}{4}.\)
三角形解的个数问题
已知两边\(a\)、\(b\)和其中一边的对角\(A\),不能确定三角形的形状,此时三角形解可能是无解、一解、两解,要分类讨论.
【例】 求满足\(a=5\),\(b=4\) ,\(A=60^{\circ}\)的三角形\(△ABC\)个数.
方法1 利用正弦定理求解
由正弦定理可得: \(\dfrac{5}{\sin 60^{\circ}}=\dfrac{4}{\sin B}\),则 \(\sin B=\dfrac{2 \sqrt{3}}{5}\),
\(\because a>b\),且\(A\)为锐角,\(\therefore B\)有一解,故三角形只有一解;
方法2 图像法
先做出角\(∠CAB=60^{\circ}\), 过点\(C\)作\(CD⊥BC\) , 此时可知\(CD=2\sqrt{3}<5\),以\(C\)为圆心,\(5\)为半径画个圆弧,由于\(b=4<a=5\),显然圆弧与射线\(AB\)交于一个点,如图可知满足题意的三角形只有一个!
面积公式
\(S_{\triangle A B C}=\dfrac{1}{2} a b \sin C=\dfrac{1}{2} b c \sin A=\dfrac{1}{2} ac\sin B\)
证明 如图,在\(△ABC\),内角\(A\) ,\(B\) ,\(C\)所对的边分别是\(a\),\(b\),\(c\),过点\(A\)作\(AD⊥BC\)交\(BC\)于点\(D\),
则 \(S_{\triangle A B C}=\dfrac{1}{2} A D \cdot B C=\dfrac{1}{2} A C \cdot \sin C \cdot B C=\dfrac{1}{2} a b \sin C\),其他类似证明可得!
基本方法
【题型1】 正弦定理解三角形
【典题1】 在\(△ABC\)中,已知\(b=2\),\(B=45^{\circ}\),\(c=\sqrt{6}\),则角\(C\)为 ( )
A.\(60^{\circ}\) \(\qquad \qquad \qquad\) B.\(30^{\circ}\)或\(150^{\circ}\) \(\qquad \qquad \qquad\)C.\(60^{\circ}\) 或\(120^{\circ}\) \(\qquad \qquad \qquad\) D.\(120^{\circ}\)
解析 由正弦定理知, \(\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\),
所以 \(\sin C=\dfrac{c \sin B}{b}=\dfrac{\sqrt{6} \times \dfrac{\sqrt{2}}{2}}{2}=\dfrac{\sqrt{3}}{2}\),
所以\(C=60^{\circ}\) 或\(120^{\circ}\) ,经检验,均符合题意.
故选:\(C\).
点拨 已知三角形的两边及其一边的对角,可用正弦定理解另一边的对角.得到 \(\sin C=\dfrac{\sqrt{3}}{2}\),则\(C=60^{\circ}\) 或\(120^{\circ}\) ,还要注意最大角\(120^{\circ}\)加上角\(B\)是否会超过\(180^{\circ}\),故还要检验.
【典题2】 在\(△ABC\)中,角\(A\) ,\(B\) ,\(C\)所对的边分别是\(a\),\(b\),\(c\),且面积为\(S\),若\(b\cos C+c\cos B=2a\cos A\), \(S=\dfrac{1}{4}\left(b^2+a^2-c^2\right)\),则角B等于( )
A.\(\dfrac{\pi}{2}\) \(\qquad \qquad \qquad\) B. \(\dfrac{5 \pi}{12}\) \(\qquad \qquad \qquad\) C. \(\dfrac{7 \pi}{12}\) \(\qquad \qquad \qquad\)D.\(\dfrac{\pi}{3}\)
解析 因为\(b\cos C+c\cos B=2a\cos A\),
由正弦定理可得,\(\sin B\cos C+\sin C\cos B=2\sin A\cos A\),
即\(\sin (B+C)=2\sin A\cos A=\sin A\),
因为\(\sin A≠0\),所以\(\cos A=\dfrac{1}{2}\),故\(A=\dfrac{1}{3} π\),
\(\because S=\dfrac{1}{4}(b^2+a^2-c^2)\),\(\dfrac{1}{2} ab\sin C=\dfrac{1}{4}×2ab×\cos C\),
\(\therefore \sin C=\cos C\),
故\(C=\dfrac{\pi}{4}\),则角 \(B=\dfrac{5 \pi}{12}\).
故选:\(B\).
点拨 对于类似"\(b\cos C+c\cos B=2a\cos A\)"即含角又含边的等式,可转化为仅含角或仅含边的等式处理;三角形的面积 \(S_{\triangle A B C}=\dfrac{1}{2} a b \sin C=\dfrac{1}{2} b c \sin A=\dfrac{1}{2} a c \sin B\)有三条,与余弦定理公式一样,一般题目中涉及哪个角就使用哪条.
【巩固练习】
1.在\(△ABC\)中,已知\(a=4\),\(b=4\sqrt{3}\),\(B=60^{\circ}\),则角\(A=\)\(\underline{\quad \quad}\) .
2.设\(△ABC\)的内角\(A\) ,\(B\) ,\(C\)所对的边分别是\(a\),\(b\),\(c\),若\(a=\sqrt{3}\),\(\sin B=\dfrac{1}{2}\),\(C=\dfrac{\pi}{6}\),则\(b=\)\(\underline{\quad \quad}\) .
3.已知\(△ABC\)的三个内角\(A\) ,\(B\) ,\(C\)所对的边分别是\(a\),\(b\),\(c\),且满足\(a\cos B+b\cos A=-\sqrt{2} c\cos A\),则\(A\)等于\(\underline{\quad \quad}\) .
4.在\(△ABC\)中,内角\(A\) ,\(B\) ,\(C\)所对的边分别是\(a\),\(b\),\(c\),且\(c=2(a\cos B-b\cos A)\),则 \(\dfrac{\tan A}{\tan B}=\)\(\underline{\quad \quad}\).
5.在\(△ABC\)中,角\(A\) ,\(B\) ,\(C\)所对的边分别是\(a\),\(b\),\(c\),且\(c=2\sqrt{3}\),\(C=\dfrac{2\pi}{3}\),\(a=4\sin B\),则\(△ABC\)的面积为\(\underline{\quad \quad}\) .
参考答案
-
答案 \(\dfrac{\pi}{6}\)
解析 \(\because a=4\),\(b=4\sqrt{3}\),\(B=60^{\circ}\) ,
\(\therefore \dfrac{a}{\sin A}=\dfrac{b}{\sin B}\),即 \(\sin A=\dfrac{a \sin B}{b}=\dfrac{4 \times \dfrac{\sqrt{3}}{2}}{4 \sqrt{3}}=\dfrac{1}{2}\),
\(\because b>a\),\(\therefore 0<A<\dfrac{\pi}{3}\),\(\therefore A=\dfrac{\pi}{6}\). -
答案 \(1\)
解析 因为\(C=\dfrac{\pi}{6}\),所以 \(B \in\left(0, \dfrac{5 \pi}{6}\right)\),
又\(\sin B=\dfrac{1}{2}\),所以\(B=\dfrac{\pi}{6}\),
所以\(A=π-(B+C)=2\dfrac{\pi}{3}\),
由正弦定理知, \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}\),
所以 \(b=\dfrac{a \sin B}{\sin A}=\dfrac{\sqrt{3} \times \dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}=1\). -
答案 \(\dfrac{3\pi}{4}\)
解析 因为\(a\cos B+b\cos A=-\sqrt{2} c\cos A\),
由正弦定理可得,\(\sin A\cos B+\sin B\cos A=-\sqrt{2} \sin C\cos A\),
即\(\sin (A+B)=\sin C=-\sqrt{2} \sin C\cos A\),
因为\(\sin C>0\),所以\(\cos A=-\dfrac{\sqrt{2}}{2}\),
因为\(A\)为三角形的内角,则\(A=\dfrac{3\pi}{4}\). -
答案 \(3\)
解析 由\(c=2(a\cos B-b\cos A)\),
根据正弦定理可得\(\sin C=2(\sin A\cos B-\sin B\cos A)\),
即\(\sin (A+B)=2(\sin A\cos B-\sin B\cos A)\),
化简可得\(3\sin B\cos A=\sin A\cos B\),
\(\therefore 3\tanB=\tanA\), \(\therefore \dfrac{\tan A}{\tan B}=3\). -
答案 \(\sqrt{3}\)
解析 \(\because \dfrac{a}{\sin A}=\dfrac{c}{\sin C}\),而\(c=2\sqrt{3}\),\(C=\dfrac{2\pi}{3}\),
\(\therefore a=\dfrac{c \sin A}{\sin C}=4 \sin A\),
又\(a=4\sin B\),\(\therefore 4\sin A=4\sin B\),\(\therefore a=b\),
而\(c^2=a^2+b^2-2ab\cos C\),即\(12=a^2+a^2+a^2\),\(\therefore a=2\),\(b=2\),
\(\therefore S_{\triangle B B C}=\dfrac{1}{2} a b \sin C=\dfrac{1}{2} \cdot 2 \cdot 2 \cdot \sin \dfrac{2}{3} \pi=\sqrt{3}\).
【题型2】 三角形个数问题
【典题1】 在\(△ABC\)中,角\(A\) ,\(B\) ,\(C\)所对的边分别是\(a\),\(b\),\(c\),若\(a=2\),\(b=2\sqrt{2}\),且三角形有两解,则角\(A\)的取值范围是( )
A. \(\left(0, \dfrac{\pi}{4}\right)\) \(\qquad \qquad \qquad\) B. \(\left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)\) \(\qquad \qquad \qquad\) C. \(\left(\dfrac{\pi}{4}, \dfrac{3 \pi}{4}\right)\) \(\qquad \qquad \qquad\)D. \(\left(\dfrac{\pi}{4}, \dfrac{\pi}{3}\right)\)
解析 方法一
由条件知\(b\sin A<a\),即\(2\sqrt{2} \sin A<2\), \(\therefore \sin A<\dfrac{\sqrt{2}}{2}\),
\(\because a<b\),\(\therefore A<B\),\(\therefore A\)为锐角,\(\therefore 0<A<\dfrac{\pi}{4}\).
方法二
如图,\(AC=2\sqrt{2}\),以\(C\)为圆心\(2\)为半径作\(⊙C\),
则\(⊙C\)上任一点(\(⊙C\)与直线\(AC\)交点除外)可为点\(B\)构成\(△ABC\),
当\(AB\)与\(⊙C\)相切时,\(AB=2\),\(∠BAC=\dfrac{\pi}{4}\),
当\(AB\)与\(⊙C\)相交时,\(∠BAC<\dfrac{\pi}{4}\),
因为三角形有两解,所以直线\(AB\)与\(⊙C\)应相交,
所以\(0<∠BAC<\dfrac{\pi}{4}\).
点拨 处理三角形个数问题,利用方法二较为简便,当然还要结合具体的已知条件.
【巩固练习】
1.(多选)已知\(△ABC\)的内角\(A\) ,\(B\) ,\(C\)所对的边分别是\(a\),\(b\),\(c\),根据下列条件解三角形,有两解的是( )
A.\(a=\sqrt{2}\),\(b=2\),\(B=120^{\circ}\) \(\qquad \qquad \qquad\) B.\(a=2\),\(b=\sqrt{3}\),\(B=45^{\circ}\) \(\qquad \qquad \qquad\)
C.\(b=3\),\(c=\sqrt{3}\),\(B=60^{\circ}\) \(\qquad \qquad \qquad\) D.\(a=2\sqrt{3}\), \(b=\sqrt{10}\),\(B=60^{\circ}\)
参考答案
- 答案 \(BD\)
解析 对于\(A\),\(a=\sqrt{2}\),\(b=2\),\(B=120^{\circ}\),\(△ABC\)是钝角三角形,只有一解;
对于\(B\),\(a=2\),\(b=\sqrt{3}\),\(B=45^{\circ}\),由正弦定理得 \(\dfrac{2}{\sin A}=\dfrac{\sqrt{3}}{\sin 45^{\circ}}\),解得 \(\sin A=\dfrac{\sqrt{2}}{\sqrt{3}}\),
又\(a>b\),且\(A∈(0,π)\),所以\(A\)有个值,三角形有两解;
对于\(C\),\(b=3\),\(c=\sqrt{3}\),\(B=60^{\circ}\),由正弦定理得 \(\dfrac{3}{\sin 60^{\circ}}=\dfrac{\sqrt{3}}{\sin C}\),解得 \(\sin C=\dfrac{1}{2}\),
由\(b>c\),所以\(B>C\),所以\(C=30^{\circ}\),三角形只有一解;
对于\(D\),\(a=2\sqrt{3}\), \(b=\sqrt{10}\),\(B=60^{\circ}\),由正弦定理得 \(\dfrac{2 \sqrt{3}}{\sin A}=\dfrac{\sqrt{10}}{\sin 60^{\circ}}\),解得 \(\sin A=\dfrac{3}{\sqrt{10}}\),
又\(b<a\),所以\(A>60^{\circ}\),所以\(A\)有两个值,三角形有两解.
故选:\(BD\).
【题型3】 解三角形的综合题
【典题1】 在\(△ABC\)中,内角\(A\) ,\(B\) ,\(C\)所对的边分别是\(a\),\(b\),\(c\),且\(2b\sin A=a\cos C+c\cos A\).
(1)求角\(A\)的大小;
(2)若\(a=1\),\(b>a\),\(\sin B=\sqrt{3} \sin C\),求\(△ABC\)的面积.
解析 (1)由正弦定理及\(2b\sin A=a\cos C+c\cos A\),
知\(2\sin B\sin A=\sin A\cos C+\sin C\cos A\),
而\(\sin A\cos C+\sin C\cos A=\sin (A+C)=\sin B\),
所以\(2\sin B\sin A=\sin B\),
因为\(\sin B≠0\),所以\(\sin A=\dfrac{1}{2}\),
又\(A∈(0,π)\),所以\(A=\dfrac{\pi}{6}\)或\(\dfrac{ 5\pi}{6}\).
(2)因为\(b>a\),所以\(A=\dfrac{\pi}{6}\),
由正弦定理及\(\sin B=\sqrt{3} \sin C\),知\(b=\sqrt{3} c\),
由余弦定理得,\(a^2=b^2+c^2-2bc\cos A\),
所以 \(1=(\sqrt{3} c)^2+c^2-2 \cdot \sqrt{3} c \cdot c \cdot \dfrac{\sqrt{3}}{2}\),解得\(c=1\),
所以\(b=\sqrt{3} c=\sqrt{3}\),
故\(△ABC\)的面积为 \(S=\dfrac{1}{2} b c \sin A=\dfrac{1}{2} \times \sqrt{3} \times 1 \times 1 \times \dfrac{1}{2}=\dfrac{\sqrt{3}}{4}\).
点拨 在三角形中,\(A+B+C=π\),则\(\sin (A+C)=\sin B\),\(\cos (A+C)=-\cos B\).
【典题2】在平面四边形\(ABCD\)中,\(∠BAC=90^{\circ}\),\(∠DAC=30^{\circ}\),\(∠DCB=150^{\circ}\) ,\(CD=1\),\(BC=2\).
(1)求证:\(\sin ^2B+\sin ^2D=1\);
(2)求\(AC\)的长.
解析 证明:(1)由题意可知,因为\(∠BAC=90^{\circ}\),\(∠DAC=30^{\circ}\),
所以\(∠BAD=∠BAC+∠DAC=90^{\circ} +30^{\circ} =120^{\circ}\) ,
由四边形\(ABCD\)的内角和定理,得\(∠BAD+∠B+∠BCD+∠D=360^{\circ}\) ,
所以\(∠B+∠D=360^{\circ} -120^{\circ} -150^{\circ} =90^{\circ}\) ,
所以\(\cos B=\cos (90^{\circ} -D)=\sin D\),
\(\because \sin ^2B+\cos ^2B=1\),\(\therefore \sin ^2B+\sin ^2D=1\);
(2)在\(△ABC\)中,由正弦定理,得 \(\dfrac{A C}{\sin B}=\dfrac{B C}{\sin \angle B A C}\),即 ①,
在\(△ACD\)中,由正弦定理,得 \(\dfrac{A C}{\sin D}=\dfrac{C D}{\sin \angle C A D}\),即 \(\dfrac{A C}{\sin D}=\dfrac{1}{\sin 30^{\circ}}=2\) ②,
由①②得, \(\dfrac{A C}{\sin B}=\dfrac{A C}{\sin D}\),即\(\sin B=\sin D\),
由(1)知,\(\sin ^2B+\sin ^2D=1\),即\(\sin ^2B+\sin ^2B=1\),解得\(\sin B=\dfrac{\sqrt{2}}{2}\),
又\(0<B<90^{\circ}\),所以\(B=45^{\circ}\) ,
在\(△ABC\)中,\(AC=BC\sin B=2×\dfrac{\sqrt{2}}{2}=\sqrt{2}\).
【巩固练习】
1.在\(△ABC\)中,内角\(A\) ,\(B\) ,\(C\)所对的边分别是\(a\),\(b\),\(c\),且 \(\cos B=\dfrac{a}{c}-\dfrac{b}{2 c}\).
(1)求角 \(C\);
(2)若\(c=2a\),求\(\sin B\).
2.已知\(△ABC\)的内角\(A\) ,\(B\) ,\(C\)所对的边分别是\(a\),\(b\),\(c\).且 \(a\operatorname{sin}(A+B)=c\operatorname{sin} \dfrac{B+C}{2}\).
(1)求\(A\); \(\qquad \qquad\) (2)若\(△ABC\)的面积为\(\sqrt{3}\),周长为\(8\),求\(a\).
3.如图所示,在\(△ABC\)中,点\(D\)为\(BC\)边上一点,且\(BD=1\),\(E\)为\(AC\)的中点,\(AE=2\), \(\cos B=\dfrac{2 \sqrt{7}}{7}\), \(\angle A D B=\dfrac{2 \pi}{3}\).
(1)求\(AD\)的长; \(\qquad \qquad\)(2)求\(△ADE\)的面积.
参考答案
-
答案 (1) \(\dfrac{\pi}{3}\) ;(2) \(\dfrac{\sqrt{3}+\sqrt{39}}{8}\)
解析 (1)因为 \(\cos B=\dfrac{a}{c}-\dfrac{b}{2 c}\),所以\(2\sin A-2\sin C\cos B-\sin B=0\),
因为\(\sin A=\sin (B+C)=\sin B\cos C+\sin C\cos B\),
所以\(2\sin B\cos C-\sin B=0\),
又\(\sin B≠0\),所以\(\cos C=\dfrac{1}{2}\),由\(0<C<π\),得\(C=\dfrac{\pi}{3}\).
(2)因为\(c=2a\),所以 \(\sin A=\dfrac{1}{2} \sin C=\dfrac{\sqrt{3}}{4}\),
因为\(c>a\),所以 \(\cos A=\dfrac{\sqrt{13}}{4}\),
所以 \(\sin B=\sin (A+C)=\sin A \cos C+\cos A \sin C=\dfrac{\sqrt{3}+\sqrt{39}}{8}\). -
答案 (1) \(\dfrac{\pi}{3}\) ;(2) \(\dfrac{13}{4}\)
解析 (1)\(△ABC\)中, \(a\operatorname{sin}(A+B)=c\operatorname{sin} \dfrac{B+C}{2}\),
\(\therefore \operatorname{asin}(\pi-C)=\operatorname{csin}\left(\dfrac{\pi}{2}-\dfrac{A}{2}\right)\), \(\therefore \operatorname{asin} C=\operatorname{ccos} \dfrac{A}{2}\);
由正弦定理得 \(\sin A \sin C=\sin C \cos \dfrac{A}{2}\),
\(\therefore \sin A=\cos \dfrac{A}{2}\),即 \(2 \sin \dfrac{A}{2} \cos \dfrac{A}{2}=\cos \dfrac{A}{2}\);
又\(A∈(0,π)\), \(\therefore \cos \dfrac{A}{2} \neq 0\), \(\therefore 2 \sin \dfrac{A}{2}=1\),即 \(\sin \dfrac{A}{2}=\dfrac{1}{2}\),
\(\therefore \dfrac{A}{2}=\dfrac{\pi}{6}\),解得\(A=\dfrac{\pi}{3}\);
(2)\(△ABC\)的面积为\(\sqrt{3}\),周长为\(8\),
\(\therefore \dfrac{1}{2} b c \sin A=\dfrac{\sqrt{3}}{4} b c=\sqrt{3}\),
\(\therefore bc=4\),…① \(a+b+c=8\),…②
由余弦定理得:\(a^2=b^2+c^2-bc\),…③
由①②③组成方程组,可得: \(\left\{\begin{array}{l} b^2+c^2+2 b c=(8-a)^2 \\ b c=4 \\ b^2+c^2=a^2+4 \end{array}\right.\),
可得:\((8-a)^2=a^2+12\),解得: \(a=\dfrac{13}{4}\). -
答案 (1) \(2\);(2) \(\dfrac{\sqrt{3}+\sqrt{39}}{4}\)
解析 (1)在\(△ABD\)中, \(\because \cos B=\dfrac{2 \sqrt{7}}{7}\),\(B∈(0,π)\),
\(\therefore \sin B=\sqrt{1-\cos ^2 B}=\sqrt{1-\left(\dfrac{2 \sqrt{7}}{7}\right)^2}=\dfrac{\sqrt{21}}{7}\),
\(\therefore \sin \angle B A D=\sin (B+\angle A D B)=\dfrac{\sqrt{21}}{7} \times \dfrac{1}{2}+\dfrac{2 \sqrt{7}}{7} \times \dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{21}}{14}\),
由正弦定理知 \(\dfrac{A D}{\sin B}=\dfrac{B D}{\sin \angle B A D}\),得 \(A D=\dfrac{B D \cdot \sin B}{\sin \angle B A D}=\dfrac{1 \times \dfrac{\sqrt{21}}{7}}{\dfrac{\sqrt{21}}{14}}=2\).
(2)由(1)知\(AD=2\),依题意得\(AC=2AE=4\),
在\(△ACD\)中,由余弦定理得\(AC^2=AD^2+DC^2-2AD\cdot DC\cdot \cos ∠ADC\),
即\(16=4+DC^2-2×2×DC\cos \dfrac{\pi}{3}\),
\(\therefore DC^2-2DC-12=0\),解得\(D C=\sqrt{13}+1\),
\(\therefore S_{\triangle A C D}=\dfrac{1}{2} A D \cdot D C \cdot \sin \angle A D C=\dfrac{1}{2} \times 2 \times(1+\sqrt{13}) \times \dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}+\sqrt{39}}{2}\),
从而 \(S_{\triangle A D E}=\dfrac{1}{2} S_{\triangle A C D}=\dfrac{\sqrt{3}+\sqrt{39}}{4}\).
分层练习
【A组---基础题】
1.在\(△ABC\)中,\(AB=\sqrt{2}\),\(BC=\sqrt{3}\),\(A=60^{\circ}\),则角\(C\)的值为( )
A.\(\dfrac{\pi}{4}\) \(\qquad \qquad \qquad \qquad\) B.\(\dfrac{3\pi}{4}\) \(\qquad \qquad \qquad \qquad\)C.\(\dfrac{\pi}{6}\) \(\qquad \qquad \qquad\qquad\) D.\(\dfrac{\pi}{4}\)或 \(\dfrac{3\pi}{4}\)
2.在\(△ABC\)中,内角\(A\) ,\(B\) ,\(C\)所对的边分别是\(a\),\(b\),\(c\),若 \(a \sin B \cos C+c \sin B \cos A=\dfrac{\sqrt{3}}{2} b\),且\(a>b\),则\(∠B=\)( )
A.\(\dfrac{\pi}{3}\) \(\qquad \qquad \qquad \qquad\) B.\(\dfrac{\pi}{6}\) \(\qquad \qquad \qquad \qquad\) C.\(\dfrac{2\pi}{3}\) \(\qquad \qquad\qquad \qquad\)D.\(\dfrac{5\pi}{6}\)
3.在\(△ABC\)中,\(a=3\),\(b=2\sqrt{6}\),\(∠B=2∠A\),则\(\sin A\)的值为( )
A. \(\dfrac{\sqrt{3}}{4}\) \(\qquad \qquad \qquad \qquad\) B.\(\dfrac{\sqrt{3}}{3}\) \(\qquad \qquad\qquad \qquad\) C.\(\dfrac{\sqrt{3}}{2}\) \(\qquad \qquad \qquad \qquad\)D.\(1\)
4.在\(△ABC\)中,角\(A\) ,\(B\) ,\(C\)所对的边分别是\(a\),\(b\),\(c\),①若\(A>B\),则\(\sin A>\sin B\);②若\(\sin 2A=\sin 2B\),则\(△ABC\)一定为等腰三角形;③若\(\sin ^2A+\sin ^2B=\sin ^2C\),则\(△ABC\)为直角三角形;④若\(△ABC\)为锐角三角形,则\(\sin A>\cos B\).以上结论中正确的有( )
A.①③ \(\qquad \qquad \qquad \qquad\) B.①④ \(\qquad \qquad \qquad \qquad\) C.①②④ \(\qquad \qquad \qquad \qquad\) D.①③④
5.(多选)在三角形\(ABC\)中,下列命题正确的有( )
A.若\(A=30^{\circ}\),\(b=4\),\(a=5\),则\(△ABC\)有两解
B.若 \(0<\tan A \cdot \tan B<1\),则\(△ABC\)一定是钝角三角形
C.若\(\cos (A-B)\cos (B-C)\cos (C-A)=1\),则\(△ABC\)一定是等边三角形
D.若\(a-b=c\cdot \cos B-c\cdot \cos A\),则\(△ABC\)的形状是等腰或直角三角形
6.在\(△ABC\)中,\(a=2\),\(b=\sqrt{3}\),\(A=2B\),则\(\cos B=\)\(\underline{\quad \quad}\) .
7.\(△ABC\)中,若\(A=60^{\circ}\),\(a=\sqrt{3}\),则 \(\dfrac{b+c}{\sin B+\sin C}\)等于\(\underline{\quad \quad}\) .
8.在\(△ABC\)中,若\(B=120^{\circ}\),\(C=15^{\circ}\),\(a=5\),则此三角形的最大边长为\(\underline{\quad \quad}\) .
9.在\(△ABC\)中,\(a\),\(b\),\(c\)分别为角\(A\) ,\(B\) ,\(C\)所对的边,且\(2a\sin B=b(\sqrt{3} \cos A-\sin A)\).
(1)求角\(A\)的大小;
(2)若\(a=1\),\(c=2\),求\(△ABC\)的面积.
10.如图,\(D\)是直角三角形\(ABC\)斜边 上一点,\(AC=\sqrt{3} DC\).
(1)若\(∠DAC=30^{\circ}\),求角\(∠ADC\)的大小;
(2)若\(BD=2DC\),且\(DC=1\),求\(AD\)的长.
11.\(△ABC\)的外接圆半径\(R=\sqrt{3}\),角\(A\) ,\(B\) ,\(C\)所对的边分别是\(a\),\(b\),\(c\),且 \(\dfrac{2 \sin A-\sin C}{\sin B}=\dfrac{\cos C}{\cos B}\)
(1)求角\(B\)和边长\(b\);
(2)求\(S_{\triangle A B C}\)的最大值及取得最大值时的\(a\),\(c\)的值,并判断此时三角形的形状.
参考答案
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答案 \(A\)
解析 由正弦定理可得, \(\dfrac{A B}{\sin C}=\dfrac{B C}{\sin A}\),故 \(\dfrac{\sqrt{2}}{\sin C}=\dfrac{\sqrt{3}}{\dfrac{\sqrt{3}}{2}}\),即 \(\sin C=\dfrac{\sqrt{2}}{2}\),
因为\(AB<BC\),故\(C<A\),且\(C\)为三角形内角,
故\(C=\dfrac{\pi}{4}\).
故选:\(A\). -
答案 \(A\)
解析 因为 \(a \sin B \cos C+c \sin B \cos A=\dfrac{\sqrt{3}}{2} b\),
由正弦定理可得, \(\sin A \sin B \cos C+\sin C \sin B \cos A=\dfrac{\sqrt{3}}{2} \sin B\),
因为\(\sin B≠0\),所以 \(\sin A \cos C+\sin C \cos A=\sin (A+C)=\dfrac{\sqrt{3}}{2}\),
所以 \(\sin B=\dfrac{\sqrt{3}}{2}\),
因为\(a>b\),则\(∠B=\dfrac{1}{3} π\).
故选:\(A\). -
答案 \(B\)
解析 \(\because a=3\),\(b=2\sqrt{6}\),\(∠B=2∠A\),
\(\therefore \sin B=\sin 2A=2\sin A\cos A\),
\(\therefore\) 由正弦定理 \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}\),
可得 \(\sin A=\dfrac{a \cdot \sin B}{b}=\dfrac{3 \times \sin 2 A}{2 \sqrt{6}}=\dfrac{6 \times \sin A \times \cos A}{2 \sqrt{6}}\),
\(\therefore \cos A=\dfrac{\sqrt{6}}{3}\),
\(\therefore \sin A=\sqrt{1-\cos ^2 A}=\dfrac{\sqrt{3}}{3}\).
故选:\(B\). -
答案 \(D\)
解析 对于①,在\(△ABC\)中,若\(∠A>∠B\),则\(a>b\),
即有\(2R\sin A>2R\sin B\),即\(\sin A>\sin B\),则①正确;
对于②,若\(\sin 2A=\sin 2B\),则\(2A=2B\)或\(2A=π-2B\),
即\(A=B\)或\(A+B=\dfrac{\pi}{2}\),
故\(△ABC\)不一定为等腰三角形,故②错误;
对于③,\(△ABC\)中,由正弦定理可得 \(\sin A=\dfrac{a}{2 R}\), \(\sin B=\dfrac{b}{2 R}\), \(\sin C=\dfrac{c}{2 R}\),
则\(\sin ^2A+\sin ^2B=\sin ^2C\)可转化为\(a^2+b^2=c^2\),
即\(△ABC\)为直角三角形,故③正确;
对于④,若\(△ABC\)为锐角三角形,\(A+B>\dfrac{\pi}{2}\),
则\(\dfrac{\pi}{2}>A>\dfrac{\pi}{2}-B\),\(\sin A>\sin (\dfrac{\pi}{2}-B)=\cos B\)即④正确.
故正确的是①③④,
故选:\(D\). -
答案 \(BCD\)
解析 由正弦定理得\(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}\),即 \(\dfrac{5}{\dfrac{1}{2}}=\dfrac{4}{\sin B}\),得 \(\sin B=\dfrac{2}{5}\),
\(\because b<a\),\(\therefore B<A\),\(\therefore B\)为锐角,
\(\therefore\)三角形\(ABC\)有一解,故选项\(A\)错误;
若 \(0<\tan A \cdot \tan B<1\),则\(\tan A>0\)且\(\tan B>0\),
所以\(A\)、\(B\)为锐角,\(\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A \tan B}>0\),
所以\(A+B\)为锐角,\(C\)为钝角,
则\(△ABC\)一定是钝角三角形,故选项\(B\)正确;
若\(\cos (A-B)\cos (B-C)\cos (C-A)=1\),
则\(\cos (A-B)=\cos (B-C)=\cos (C-A)=1\),
则\(A-B=B-C=C-A=0\),所以\(A=B=C\),
则\(△ABC\)一定是等边三角形,故选项\(C\)正确;
若\(a-b=c \cdot \cos B-c \cdot \cos A\),
由正弦定理得\(\sin A-\sin B=\sin C\cdot \cos B-\sin C\cdot \cos A\),
即\(\sin (B+C)-\sin (A+C)=\sin C\cos B-\sin C\cos A\),整理得:\((\sin B-\sin A)\cdot \cos C=0\),
所以\(\cos C=0\)或\(\sin B-\sin A=0\),即\(C=\dfrac{\pi}{2}\) 或\(B=A\),
故\(△ABC\)是等腰或直角三角形,
所以选项\(D\)正确.
故选:\(BCD\). -
答案 \(\dfrac{\sqrt{3}}{3}\)
解析 由正弦定理得, \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}\) ,所以 \(\dfrac{2}{\sin 2 B}=\dfrac{\sqrt{3}}{\sin B}\) ,
所以 \(\cos B=\dfrac{\sqrt{3}}{3}\). -
答案 \(2\)
解析 由条件利用正弦定理可得 \(\dfrac{a}{\sin A}=2 R=\dfrac{\sqrt{3}}{\sin 60^{\circ}}=2\),
\(\therefore \dfrac{b+c}{\sin B+\sin C}=\dfrac{2 R \sin B+2 R \sin C}{\sin B+\sin C}=2 R=2\). -
答案 \(\dfrac{5 \sqrt{6}}{2}\)
解析 因为\(A+B+C=180^{\circ}\) ,所以\(A=180^{\circ} -(B+C)=45^{\circ}\),
所以最大的边为\(b\),
由正弦定理知, \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}\),
所以 \(\dfrac{5}{\sin 45^{\circ}}=\dfrac{b}{\sin 120^{\circ}}\),解得 \(b=\dfrac{5 \sqrt{6}}{2}\). -
答案 (1) \(\dfrac{\pi}{6}\);(2) \(\dfrac{\sqrt{3}}{2}\).
解析 (1)由正弦定理得\(2\sin A\sin B=\sin B(\sqrt{3} \cos A-\sin A)\),
即\(2\sin A\sin B=\sqrt{3} \sin B\cos A-\sin B\sin A\),
化简得\(\sqrt{3} \sin A\sin B=\sin B\cos A\),
又\(\sin B≠0\),故\(\sqrt{3} \sin A=\cos A\),即 \(\tan A=\dfrac{\sqrt{3}}{3}\),
又\(A∈(0,π)\),故\(A=\dfrac{\pi}{6}\);
(2)由余弦定理得\(a^2=b^2+c^2-2bc\cos A\),
即\(1=b^2+4-2\sqrt{3} b\),解得\(b=\sqrt{3}\),
故 的面积为 \(\dfrac{1}{2} b c \sin A=\dfrac{\sqrt{3}}{2}\). -
答案 (1) \(120^{\circ}\);(2)\(\sqrt{2}\)
解析 (1)在\(△ADC\)中,由正弦定理得 \(\dfrac{A C}{\sin \angle A D C}=\dfrac{D C}{\sin \angle D A C}\),
所以, \(\sin \angle A D C=\dfrac{A C \cdot \sin \angle D A C}{D C}=\sqrt{3} \times \dfrac{1}{2}=\dfrac{\sqrt{3}}{2}\).
又\(∠ADC=B+∠BAD=B+(90^{\circ} -∠DAC)=B+60^{\circ} >60^{\circ}\)
所以,\(∠ADC=120^{\circ}\).
(2)由\(BD=2DC\),且\(DC=1\)知:\(BC=3\),\(AC=\sqrt{3}\),
所以,直角三角形\(ABC\)中, \(\cos C=\dfrac{A C}{B C}=\dfrac{\sqrt{3}}{3}\).
在\(△ADC\)中,由余弦定理得
\(A D^2=A C^2+D C^2-2 A C \cdot D C \cos C\)\(=(\sqrt{3})^2+1-2 \sqrt{3} \times 1 \times \dfrac{\sqrt{3}}{3}=2\),
所以,\(AD=\sqrt{2}\). -
答案 (1)\(3\);(2)\(a=b=c=3\),可得\(△ABC\)是等边三角形.
解析 (1) \(\because \dfrac{2 \sin A-\sin C}{\sin B}=\dfrac{\cos C}{\cos B}\),
\(\therefore 2\sin A\cos B-\sin C\cos B=\sin B\cos C\),
可得\(2\sin A\cos B=\sin B\cos C+\cos B\sin C=\sin (B+C)\),
\(\because\) 在\(△ABC\)中,\(\sin (B+C)=\sin (π-A)=\sin A>0\),
\(\therefore 2\sin A\cos B=\sin A\),可得\(\cos B=\dfrac{1}{2}\).
又\(\because B∈(0,π)\),\(\therefore B=\dfrac{\pi}{3}\),
由正弦定理 \(\dfrac{b}{\sin B}=2 R\),可得 \(b=2 R \sin B=2 \sqrt{3} \cdot \sin \dfrac{\pi}{3}=3\);
(2)\(\because b=3\),\(\cos B=\dfrac{1}{2}\),
\(\therefore\)由余弦定理\(b^2=a^2+c^2-2ac\cos B\),得\(a^2+c^2-ac=9\),
因此,\(ac+9=a^2+c^2≥2ac\),可得\(ac≤9\),当且仅当\(a=c\)时等号成立,
\(\because S_{\triangle A B C}=\dfrac{1}{2} a c \sin B=\dfrac{\sqrt{3}}{4} a c\), \(\therefore S_{\triangle A B C} \leq \dfrac{\sqrt{3}}{4} \times 9=\dfrac{9 \sqrt{3}}{4}\),
由此可得:当且仅当\(a=c\)时, \(S_{\triangle A B C}\)有最大值 \(\dfrac{9 \sqrt{3}}{4}\),
此时\(a=b=c=3\),可得\(△ABC\)是等边三角形.
【B组---提高题】
1.在\(△ABC\)中,内角\(A\) ,\(B\) ,\(C\)所对的边分别是\(a\),\(b\),\(c\),若 \(\dfrac{a}{2 \cos A}=\dfrac{b}{3 \cos B}=\dfrac{c}{5 \cos C}\),则\(∠B\)的大小是( )
A. \(\dfrac{\pi}{12}\) \(\qquad \qquad \qquad\) B.\(\dfrac{\pi}{6}\) \(\qquad \qquad \qquad\)C.\(\dfrac{\pi}{4}\) \(\qquad \qquad \qquad\) D.\(\dfrac{\pi}{3}\)
2.在\(△ABC\)中,角\(A\) ,\(B\) ,\(C\)所对的边分别是\(a\),\(b\),\(c\),若\(A=3B\),则\(\dfrac{a}{b}\)的取值范围是( )
A.\((0,3)\) \(\qquad \qquad \qquad\) B.\((1,3)\) \(\qquad \qquad \qquad\) C.\((0,1]\) \(\qquad \qquad \qquad\) D.\((1,2]\)
3.如图,在平面四边形\(ABCD\)中,\(AD=1\),\(AB=2\),\(BC=CD=DB\),设\(∠DAB=θ\).
(1)若\(θ=\dfrac{2\pi}{3}\),求\(\sin ∠ADB\)的值;
(2)用\(θ\)表示四边形\(ABCD\)的面积\(S(θ)\),并求\(S(θ)\)的最大值.
参考答案
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答案 \(D\)
解析 由正弦定理可知,\(a=2R\sin A\),\(b=2R\sin B\),\(c=2R\sin C\),(\(R\)为三角形外接圆半径),
因为 \(\dfrac{a}{2 \cos A}=\dfrac{b}{3 \cos B}=\dfrac{c}{5 \cos C}\),
所以, \(\dfrac{\sin A}{2 \cos A}=\dfrac{\sin B}{3 \cos B}=\dfrac{\sin C}{5 \cos C}\),且\(A\) ,\(B\) ,\(C\)都为锐角,
所以 \(\dfrac{1}{2} \tan A=\dfrac{1}{3} \tan B=\dfrac{1}{5} \tan C\),
所以 \(-\tan B=\tan (A+C)=\dfrac{\tan A+\tan C}{1-\tan A \tan C}=\dfrac{\dfrac{2 \tan B}{3}+\dfrac{5 \tan B}{3}}{1-\dfrac{2 \tan B}{3} \cdot \dfrac{5 \tan B}{3}}\),
整理可得,\(\tan ^2B=3\),
故\(\tan B=\sqrt{3}\),\(B=\dfrac{1}{3} π\).
故选:\(D\). -
答案 \(B\)
解析 \(\because A=3B\),
\(\therefore\) 由正弦定理得 \(\dfrac{a}{b}=\dfrac{\sin A}{\sin B}=\dfrac{\sin 3 B}{\sin B}=\dfrac{\sin B \cos 2 B+\cos B \sin 2 B}{\sin B}\)\(=\cos 2 B+2 \cos ^2 B=2 \cos 2 B+1\),
\(\because B+A<180^{\circ}\),即\(4B<180^{\circ}\),
\(\therefore 0<B<45^{\circ}\),即\(0<2B<90^{\circ}\),
\(\therefore 0<\cos 2B<1\),即\(1<2\cos 2B+1<3\),
则\(\dfrac{a}{b}\)的取值范围为\((1,3)\).
故选:\(B\). -
答案 (1) \(\dfrac{\sqrt{21}}{7}\) ;(2) \(S(\theta)=2 \sin \left(\theta-\dfrac{\pi}{3}\right)+\dfrac{5 \sqrt{3}}{4}\),最大值为 \(2+\dfrac{5 \sqrt{3}}{4}\).
解析 (1)在\(△ABC\)中,由余弦定理知\(BD^2=AD^2+AB^2-2AD\cdot AB\cos ∠BAD\),
由已知\(AD=1\),\(AB=2\),\(∠DAB=\dfrac{2}{3} π\),
代入上式得:\(B D^2=1+4-2 \times 1 \times 2 \times\left(-\dfrac{1}{2}\right)=7\),即\(BD=\sqrt{7}\),
又由正弦定理得: \(\dfrac{A B}{\sin \angle A D B}=\dfrac{B D}{\sin \angle D A B}\)
即: \(\dfrac{2}{\sin \angle A D B}=\dfrac{\sqrt{7}}{\sin \dfrac{2}{3} \pi}\),解得: \(\sin \angle A D B=\dfrac{\sqrt{21}}{7}\).
(2)在\(△ABC\)中,由余弦定理知\(BD^2=1+4-2×1×2×\cos θ=5-4\cos θ\),
\(\because △BDC\)为等边三角形,且边长为\(5-4\cos θ\).
故 \(S(\theta)=S_{\triangle A B D}+S_{\triangle B D C}=\dfrac{1}{2} A D \cdot A B \cdot \sin \theta+\dfrac{1}{2} B D \cdot C D \cdot \sin 60^{\circ}\)
\(=\dfrac{1}{2} \cdot 1 \cdot 2 \sin \theta+\dfrac{1}{2} \cdot B D^2 \cdot \sin 60^{\circ}=\sin \theta+\dfrac{\sqrt{3}(5-4 \cos \theta)}{4}\)
\(=\sin \theta-\sqrt{3} \cos \theta+\dfrac{5 \sqrt{3}}{4}=2 \sin \left(\theta-\dfrac{\pi}{3}\right)+\dfrac{5 \sqrt{3}}{4}\).
\(\because 0<θ<π\),\(\therefore -\dfrac{\pi}{3}<θ-\dfrac{\pi}{3}<\dfrac{2\pi}{3}\),
故当\(θ-\dfrac{\pi}{3}=\dfrac{\pi}{2}\)时,\(S(θ)\)取得最大值为\(2+\dfrac{5 \sqrt{3}}{4}\).
【C组---拓展题】
1.已知在锐角\(△ABC\)中,角\(A\) ,\(B\) ,\(C\)所对的边分别是\(a\),\(b\),\(c\),若\(2b\cos (A+B)=c\cos (A+C)\),则 \(\dfrac{1}{\tan A}+\dfrac{1}{\tan B}+\dfrac{1}{\tan C}\)的最小值为( )
A. \(\dfrac{2 \sqrt{7}}{3}\) \(\qquad \qquad \qquad\) B.\(\sqrt{5}\) \(\qquad \qquad \qquad\)C. \(\dfrac{\sqrt{7}}{3}\) \(\qquad \qquad \qquad\) D.\(2\sqrt{5}\)
2.在\(△ABC\)中,已知 \(2 \cos ^2 \dfrac{A}{2}=\dfrac{\sqrt{3}}{3} \sin A\),若\(a=2\sqrt{3}\),则\(△ABC\)周长的取值范围为\(\underline{\quad \quad}\).
3.边长为\(1\)的正方形\(ABCD\)的边\(BC\)上有一点\(P\),边\(CD\)上有一点\(Q\).满足\(△CPQ\)的周长为\(2\).
(1)求\(∠QAP\)的大小;(2)求\(△APQ\)面积的最小值.
参考答案
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答案 \(A\)
解析 因为\(2b\cos (A+B)=c\cos (A+C)\),得\(2b\cos C=c\cos B\),
由正弦定理得\(2\sin B\cos C=\sin C\cos B\),
所以\(2\tan B=\tan C\),
又因为\(A+B+C=π\),
所以 \(\tan A=\tan [\pi-(B+C)]=-\tan (B+C)\)
\(=-\dfrac{\tan B+\tan C}{1-\tan B \tan C}=\dfrac{-3 \tan B}{1-2 \tan ^2 B}\),
所以 \(\dfrac{1}{\tan A}+\dfrac{1}{\tan B}+\dfrac{1}{\tan C}=\dfrac{1-2 \tan ^2 B}{-3 \tan B}+\dfrac{1}{\tan B}+\dfrac{1}{2 \tan B}\)\(=\dfrac{2 \tan ^2 B-1}{3 \tan B}+\dfrac{3}{2 \tan B}=\dfrac{9+4 \tan ^2 B-2}{6 \tan B}\)
\(=\dfrac{4 \tan ^2 B+7}{6 \tan B}=\dfrac{2}{3} \tan B+\dfrac{7}{6 \tan B}\)
\(\geq 2 \sqrt{\dfrac{2}{3} \tan B \times \dfrac{7}{6 \tan B}}=\dfrac{2 \sqrt{7}}{3}\),
(当且仅当 \(\dfrac{2}{3} \tan B=\dfrac{7}{6 \tan B}\),即 \(\tan B=\dfrac{\sqrt{7}}{2}\),取“=”).
所以\(\dfrac{1}{\tan A}+\dfrac{1}{\tan B}+\dfrac{1}{\tan C}\)的最小值为 \(\dfrac{2 \sqrt{7}}{3}\).
故选:\(A\). -
答案 \((4 \sqrt{3}, 4+2 \sqrt{3}]\)
解析 由 \(12 \cos ^2 \dfrac{A}{2}=\dfrac{\sqrt{3}}{3} \sin A\)得 \(1+\cos A=\dfrac{\sqrt{3}}{3} \sin A\),
即 \(\dfrac{\sqrt{3}}{3} \sin A-\cos A=1\),
即 \(\dfrac{2 \sqrt{3}}{3} \sin \left(A-\dfrac{\pi}{3}\right)=1\),即 \(\sin \left(\mathrm{A}-\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\),
\(\because 0<A<π\),\(\therefore -\dfrac{\pi}{3}<A-\dfrac{\pi}{3}<\dfrac{2\pi}{3}\),
即\(A-\dfrac{\pi}{3}=\dfrac{\pi}{3}\),即\(A=\dfrac{2\pi}{3}\),
\(\because \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=\dfrac{2 \sqrt{3}}{\dfrac{\sqrt{3}}{2}}=4\),\(\therefore b=4\sin B\),\(c=4\sin C\),
则 \(b+c=4 \sin B+4 \sin C=4 \sin B+4 \sin \left(\dfrac{\pi}{3}-B\right)\)
\(=4 \sin B+4\left(\dfrac{\sqrt{3}}{2} \cos B-\dfrac{1}{2} \sin B\right)\)
\(=2 \sin B+2 \sqrt{3} \cos B=4 \sin \left(B+\dfrac{\pi}{3}\right)\),
\(\because 0<B<\dfrac{\pi}{3}\),\(\therefore \dfrac{\pi}{3}<B+\dfrac{\pi}{3}<\dfrac{2\pi}{3}\),
则 \(\dfrac{\sqrt{3}}{2}<\sin \left(B+\dfrac{\pi}{3}\right) \leq 1\),即 \(2 \sqrt{3}<4 \sin \left(B+\dfrac{\pi}{3}\right) \leq 4\),
则\(2\sqrt{3}<b+c≤4\),
则\(4\sqrt{3}<a+b+c≤4+2\sqrt{3}\),
即三角形的周长的范围是\((4\sqrt{3},4+2\sqrt{3}]\),
故答案为:\((4\sqrt{3},4+2\sqrt{3}]\). -
答案 (1)\(45^{\circ}\);(2)\(\sqrt{2}-1\)
解析 (1)\(△CPQ\)的周长为\(2\),可得\(PQ=BP+DQ\),
设\(∠PAB=α\),\(∠DAQ=β\),
则\(PB=\tan α\),\(QD=\tan β\),且\(PC=1-\tan α\),\(QC=1-\tan β\),\(PQ=\tan α+\tan β\),
由勾股定理可得,\((1-\tan α)^2+(1-\tan β)^2=(\tan α+\tan β)^2\),
展开整理可得,\(2-2\tan α-2\tan β=2\tan α\cdot \tan β\),
变形可得 \(\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=1\),即\(\tan (α+β)=1\),
因为\(α+β\)为锐角,\(α+β=45^{\circ}\),\(∠QAP=90^{\circ}-45^{\circ}=45^{\circ}\),
(2) \(S_{\triangle A P Q}=\dfrac{1}{2} A Q \cdot A P \cdot \sin 45^{\circ}=\dfrac{1}{2 \sqrt{2} \cos \alpha \cos \beta}\),
又\(2\sqrt{2} \cos α\cos β=2\sqrt{2} \cos α\cos (45^{\circ}-α)=2\cos 2α+2\cos α\sin α\),
\(2\sqrt{2} \cos α\cos β=1+\cos 2α+\sin 2α=1+\sqrt{2} \cos (45^{\circ}-2α)\),
当\(α=22.5^{\circ}\)时,上式有最大值\(1+\sqrt{2}\),
此时 \(S_{\triangle A P Q}=\dfrac{1}{1+\sqrt{2} \cos \left(45^{\circ}-2 \alpha\right)}\)有最小值\(\sqrt{2}-1\).