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[ 【基础过关系列】高一数学同步精品讲义与分层练习(人教A版2019)]
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必修第二册同步巩固,难度2颗星!
基础知识
解三角形
一般地,三角形的三个角\(A\),\(B\),\(C\)和它们的对边\(a\),\(b\),\(c\)叫做三角形的元素。已知三角形的几个元素求其他元素的过程叫做解三角形.
余弦定理
(1) 内容
三角形中任何一边的平方,等于其他两边平方的和减去这两边与它们夹角的余弦的积的两倍.
即\(a^2=b^2+c^2-2bc\cos A\) ,\(b^2=a^2+c^2-2ac\cos B\),\(c^2=a^2+b^2-2ab\cos C\).
证明 因为 \(|\overrightarrow{B C}|^2=\overrightarrow{B C}^2=(\overrightarrow{A C}-\overrightarrow{A B})^2=\overrightarrow{A C}^2-2 \overrightarrow{A C} \cdot \overrightarrow{A B}+\overrightarrow{A B}^2\)
\(=\overrightarrow{A C}^2-2|\overrightarrow{A C}|\cdot |\overrightarrow{AB} | \cos A+\overrightarrow{AB}^2\)
所以\(a^2=b^2+c^2-2bc\cos A\),
同理可得\(b^2=a^2+c^2-2ac\cos B\),\(c^2=a^2+b^2-2ab\cos C\).
(2) 变形
\(\cos A=\dfrac{b^2+c^2-a^2}{2 b c}\) , \(\cos B=\dfrac{a^2+c^2-b^2}{2 a c}\) , \(\cos C=\dfrac{a^2+b^2-c^2}{2 a b}\)
(3)利用余弦定理可以解决下列两类三角形的问题
① 已知三边,可求三个角;
【例】 在\(△ABC\)中,若\(a=4\) , \(b=3\), \(c=\sqrt{13}\),则角\(C=\)\(\underline{\quad \quad}\) .
解 \(\cos C=\dfrac{a^2+b^2-c^2}{2 a b}=\dfrac{16+9-13}{24}=\dfrac{1}{2} \Rightarrow C=\dfrac{\pi}{3}\).
② 已知两边和一角,求第三边和其他两个角.
【例1】 在\(△ABC\)中,\(A=30°\) , \(b=\sqrt{3}\),\(c=1\),则\(a=\)\(\underline{\quad \quad}\) .(角\(A\)为两边的夹角)
解 \(a^2=b^2+c^2-2bc\cos A=3+1-3=1⇒a=1\).
【例2】 在\(△ABC\)中,\(A=30°\) , \(b=3\sqrt{3}\),\(a=3\), 则边\(c=\)\(\underline{\quad \quad}\). (角\(A\)不为两边的夹角)
解 \(a^2=b^2+c^2-2bc\cos A⇒9=27+c^2-9c⇒c=3\)或\(c=6\).
三角形类型的判断
\(\angle A=\dfrac{\pi}{2} \Rightarrow b^2+c^2=a^2\);
\(\angle A>\dfrac{\pi}{2} \Rightarrow \cos A=\dfrac{b^2+c^2-a^2}{2 b c}<0 \Rightarrow b^2+c^2<a^2\);
\(\angle A<\dfrac{\pi}{2} \Rightarrow \cos A=\dfrac{b^2+c^2-a^2}{2 b c}>0 \Rightarrow b^2+c^2>a^2\).
射影定理
\(a=c \cdot \cos B+b \cdot \cos C\) ,\(b=a \cdot \cos C+c \cdot \cos A\),\(c=b \cdot \cos A+a \cdot \cos B\)
基本方法
【题型1】 余弦定理解三角形
【典题1】 若\(△ABC\)的三边长分别为\(3\)、\(6\)、\(7\),则该三角形最大角的余弦值为 \(\underline{\quad \quad}\).
解析 \(\because△ABC\)的三边长分别为\(3\)、\(6\)、\(7\),
\(\therefore\) 该三角形最大角的余弦值为 \(\dfrac{3^2+6^2-7^2}{2 \times 3 \times 6}=-\dfrac{1}{9}\).
点拨 三角形中大角对大边;已知三角形的三边可用余弦定理求三内角.
【典题2】\(△ABC\)的内角\(A\),\(B\),\(C\)和它们的对边\(a\),\(b\),\(c\),已知\(a=\sqrt{5}\),\(c=2\), \(\cos A=\dfrac{2}{3}\),则\(b=\)\(\underline{\quad \quad}\).
解析 因为\(a=\sqrt{5}\),\(c=2\),\(\cos A=\dfrac{2}{3}\),
所以由余弦定理可得: \(\cos A=\dfrac{b^2+c^2-a^2}{2 b c}\),即 \(\dfrac{2}{3}=\dfrac{b^2+4-5}{4 b}\),
整理可得:\(3b^2-8b-3=0\),解得\(b=3\)或 \(-\dfrac{1}{3}\)(舍去),
所以\(b=3\).
点拨 已知三角形的两边与一角,可用余弦定理求第三边.余弦定理有三条,那一般题中涉及哪个角就用对应的余弦定理公式.
【巩固练习】
1.在\(△ABC\)中,\(A=30°\),\(b=\sqrt{3}\),\(c=1\),则\(a=\)( )
A.\(2\) \(\qquad \qquad \qquad\) B.\(\sqrt{3}\) \(\qquad \qquad \qquad\) C.\(\sqrt{2}\) \(\qquad \qquad \qquad\) D.\(1\)
2.在\(△ABC\)中,内角\(A\),\(B\),\(C\)和它们的对边\(a\),\(b\),\(c\).若\(a=3\),\(A=30°\),\(b=3\sqrt{3}\),则\(c\)值为( )
A.\(3\) \(\qquad \qquad \qquad\) B.\(3\)或\(6\) \(\qquad \qquad \qquad\) C.\(\sqrt{3}\) \(\qquad \qquad \qquad\) D.\(\sqrt{3}\)或\(6\)
3.在\(△ABC\)中,已知角\(A\),\(B\),\(C\)和它们的对边\(a\),\(b\),\(c\),\(a=1\),\(b=\sqrt{2}\),\(C=45^∘\),则边\(c\)等于\(\underline{\quad \quad}\) .
4.在\(△ABC\)中,若\(ac=8\),\(a+c=7\), \(B=\dfrac{\pi}{3}\),则\(b=\)\(\underline{\quad \quad}\).
参考答案
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答案 \(D\)
解析 因为\(A=30°\),\(b=\sqrt{3}\),\(c=1\),
\(\therefore a^2=b^2+c^2-2 b \cos A=\sqrt{3}^2+1^2-2 \times \sqrt{3} \times 1 \times \cos 30^{\circ}=1\),
故\(a=1\).
故选:\(D\). -
答案 \(B\)
解析 由余弦定理可得\(a^2=b^2+c^2-2bc\cos A\),
即\(9=27+c^2-9c\),即\(c^2-9c+18=0\),解得\(c=3\)或\(c=6\),
故选:\(B\). -
答案 \(1\)
解析 由余弦定理得, \(c=\sqrt{a^2+b^2-2 a b \cos C}=\sqrt{1+2-2 \times 1 \times \sqrt{2} \times \dfrac{\sqrt{2}}{2}}=1\). -
答案 \(5\)
解析 由余弦定理知,\(b^2=a^2+c^2-2ac\cos B=(a+c)^2-2ac-2ac \cos B\)
\(=49-2 \times 8-2 \times 8 \times \dfrac{1}{2}=25\),所以\(b=5\).
【题型2】 余弦定理的运用
【典题1】 已知\(△ABC\)中,角\(A\),\(B\),\(C\)和它们的对边\(a\),\(b\),\(c\).若\(2b\cos C=a+2c\cos B\),\(b=\sqrt{2} c\),则\(\cos C=\)\(\underline{\quad \quad}\).
解析 由余弦定理及\(2b\cos C=a+2c\cos B\)知, \(2 b \cdot \dfrac{a^2+b^2-c^2}{2 a b}=a+2 c \cdot \dfrac{a^2+c^2-b^2}{2 a c}\),
化简可得\(a^2=2(b^2-c^2 )\),
因为\(b=\sqrt{2} c\),所以\(a^2=2(2c^2-c^2 )=2c^2\),即\(a=\sqrt{2} c\),
由余弦定理知, \(\cos C=\dfrac{a^2+b^2-c^2}{2 a b}=\dfrac{2 c^2+2 c^2-c^2}{2 \cdot \sqrt{2} c \cdot \sqrt{2} c}=\dfrac{3}{4}\).
点拨 遇到类似"\(2b\cos C=a+2c\cos B\)"含角含边的等式,可化为仅含角或仅含边的等式.
【典题2】在\(△ABC\)中,角\(A\),\(B\),\(C\)和它们的对边\(a\),\(b\),\(c\),已知\(a=4\),\(b=5\),\(c=6\),则\(BC\)边上的中线长\(AD=\)\(\underline{\quad \quad}\).
解析 因为\(a=4\),\(b=5\),\(c=6\),
所以 \(\cos B=\dfrac{a^2+c^2-b^2}{2 a c}=\dfrac{16+36-25}{2 \times 4 \times 6}=\dfrac{9}{16}\),
又\(AD\)为\(BC\)边上的中线长,
所以 \(B D=\dfrac{a}{2}=2\),
在\(△ABD\)中,由余弦定理可得 \(A D^2=c^2+\left(\dfrac{a}{2}\right)^2-2 \cdot c \cdot \dfrac{a}{2} \cdot \cos B=36+4-2 \times 6 \times 2 \times \dfrac{9}{16}=\dfrac{53}{2}\),
可得 \(A D=\dfrac{\sqrt{106}}{2}\).
点拨 对于类似本题图象含有多个三角形的问题,一是尽量去思考在每个三角形中哪些角哪些边可求尽量先确定;二是注意公角或公边的三角形间条件转换.
【巩固练习】
1.在\(△ABC\)中,角\(A\),\(B\),\(C\)和它们的对边\(a\),\(b\),\(c\).若 \(b^2+c^2-a^2=\dfrac{6}{5} b c\),则\(\sin A\)的值为\(\underline{\quad \quad}\).
2.在\(△ABC\)中,内角\(A\),\(B\),\(C\)和它们的对边\(a\),\(b\),\(c\),若\(2a^2=2b^2+bc\), \(\cos A=\dfrac{1}{4}\),则 \(\dfrac{b}{c}=\) \(\underline{\quad \quad}\) .
3.设\(△ABC\)的内角\(A\),\(B\),\(C\)和它们的对边\(a\),\(b\),\(c\).若\(b^2=ac\),\(a+c=4\), \(\overrightarrow{B A} \cdot \overrightarrow{B C}=3\),则\(\cos B=\)\(\underline{\quad \quad}\).
4.在钝角三角形\(ABC\)中,\(a=1\),\(b=2\),则边\(c\)的取值范围是\(\underline{\quad \quad}\).
5.在\(△ABC\)中,\(D\)是\(AB\)边上一点,\(AD=2DB\),\(DC⊥AC\),\(DC=\sqrt{3}\), \(B C=\sqrt{7}\),则\(AB=\)\(\underline{\quad \quad}\).
参考答案
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答案 \(\dfrac{4}{5}\)
解析 \(\because b^2+c^2-a^2=\dfrac{6}{5} b c\),
又由余弦定理可得,\(b^2+c^2-a^2=2bc⋅\cos A\),
\(\therefore \dfrac{6}{5} b c=2 b c \cdot \cos A\),即 \(\cos A=\dfrac{3}{5}\),
\(\therefore \sin A=\sqrt{1-\cos ^2 A}=\dfrac{4}{5}\). -
答案 \(1\)
解析 在\(△ABC\)中,\(2a^2=2b^2+bc\),整理得 \(b^2=a^2-\dfrac{1}{2} b c\),
由余弦定理: \(\cos A=\dfrac{b^2+c^2-a^2}{2 b c}=\dfrac{c^2-\dfrac{1}{2} b c}{2 b c}=\dfrac{2 c-b}{4 b}=\dfrac{1}{4}\),
整理得 \(\dfrac{b}{c}=1\). -
答案 \(\dfrac{9}{10}\)
解析 由\(\overrightarrow{B A} \cdot \overrightarrow{B C}=3\),得\(ca\cos B=3\),
结合余弦定理可得 \(a c \cdot \dfrac{a^2+c^2-b^2}{2 a c}=3\),
\(\therefore a^2+c^2-b^2=6\).\(\therefore (a+c)^2-2ac-b^2=6\),
\(\therefore 3b^2=10\), \(\therefore b^2=\dfrac{10}{3}\),
\(\therefore \cos B=\dfrac{9}{10}\) . -
答案 \(1<c<\sqrt{3}\)或\(\sqrt{5}<c<3\).
解析 ①\(∵\)当\(∠C\)是钝角时,有\(∠C>90°\),
\(\therefore c>\sqrt{a^2+b^2}=\sqrt{5}\),
又\(a+b>c\),可得\(c<1+2=3\),
\(\therefore\) 可得边\(c\)的取值范围是\((\sqrt{5},3)\);
②当\(∠B\)是钝角时,有\(∠B>90°\),
\(\therefore b^2>a^2+c^2\),可得\(4>1+c^2\),解得\(c<\sqrt{3}\),
又\(c>b-a=1\),
\(\therefore 1<c<\sqrt{3}\),
综上,边\(c\)的取值范围是\(1<c<\sqrt{3}\)或\(\sqrt{5}<c<3\). -
答案 \(1\)
解析 如图,设\(BD=x\),则由余弦定理可得, \(\cos A=\dfrac{\sqrt{4 x^2-3}}{2 x}\),
又由余弦定理可得, \(7=B C^2=9 x^2+\left(4 x^2-3\right)-2 \cdot 3 x \cdot \sqrt{4 x^2-3} \cos A\)
\(=13 x^2-6 x \cdot \sqrt{4 x^2-3} \times \dfrac{\sqrt{4 x^2-3}}{2 x}-3\),
即\(7=6+x^2\),解得\(x=1\),
\(\therefore AB=3\).
故答案为:\(1\).
分层练习
【A组---基础题】
1.已知\(a\),\(b\),\(c\)分别为\(△ABC\)内角\(A\),\(B\),\(C\)的对边,若\(c=3\),\(b=\sqrt{3}\),\(B=30°\),则\(a=\)( )
A.\(3\) \(\qquad \qquad \qquad\) B.\(2\sqrt{3}\) \(\qquad \qquad \qquad\) C.\(\sqrt{3}\)或\(2\sqrt{3}\) \(\qquad \qquad \qquad\) D.\(3\)或\(2\sqrt{3}\)
2.在\(△ABC\)中,角\(A\),\(B\),\(C\)和它们的对边\(a\),\(b\),\(c\),若\(a=4\),\(b=3\), \(c=\sqrt{13}\),则\(C=\)( )
A.\(30°\) \(\qquad \qquad \qquad\) B.\(45°\) \(\qquad \qquad \qquad\) C.\(60°\) \(\qquad \qquad \qquad\) D.\(120°\)
3.(多选)已知\(△ABC\)中,角\(A\),\(B\),\(C\)和它们的对边\(a\),\(b\),\(c\),且满足\(B=\dfrac{\pi}{3}\),\(a+c=\sqrt{3} b\),则 \(\dfrac{a}{c}=\)( )
A.\(2\) \(\qquad \qquad \qquad\) B.\(3\) \(\qquad \qquad \qquad\) C. \(\dfrac{1}{2}\) \(\qquad \qquad \qquad\) D. \(\dfrac{1}{3}\)
4.在\(△ABC\)中,\(a^2=b^2+c^2+bc\),则\(∠A=\)\(\underline{\quad \quad}\) .
5.在\(△ABC\)中,若\(a=2\),\(b=4\), \(\cos C=\dfrac{1}{4}\),则\(△ABC\)的周长等于\(\underline{\quad \quad}\).
6.已知\(△ABC\)中,角\(A\),\(B\),\(C\)和它们的对边\(a\),\(b\),\(c\),且满足 \(a c \cos B=a^2-b^2+\dfrac{1}{2} b c\),则\(A=\)\(\underline{\quad \quad}\).
7.在\(△ABC\)中,角\(A\),\(B\),\(C\)和它们的对边\(a\),\(b\),\(c\).已知 \(C=\dfrac{\pi}{3}\), \(b^2-c^2=\dfrac{1}{2} a^2\),则\(\cos A=\)\(\underline{\quad \quad}\).
8.在\(△ABC\)中,内角\(A\),\(B\),\(C\)和它们的对边\(a\),\(b\),\(c\).已知\(c=2\),\(b=1\), \(\cos C=\dfrac{1}{4}\).则\(△ABC\)的中线\(AD\)的长为\(\underline{\quad \quad}\).
9.如图所示,在平面四边形\(ABCD\)中, \(A B=\sqrt{10}\),\(BC=3\),\(AC=5\),\(CD=2\sqrt{2}\),\(∠BCD=135^∘\).
(1)求\(\sin∠ACB\); \(\qquad \qquad\)(2)求\(AD\)的长.
参考答案
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答案 \(C\)
解析 \(\because c=3\),\(b=\sqrt{3}\),\(B=30°\),
\(\therefore\) 由余弦定理可得:\(b^2=a^2+c^2-2ac\cos B\),
可得 \(3=a^2+3^2-2 \times a \times 3 \times \dfrac{\sqrt{3}}{2},\),可得\(a^2-3\sqrt{3} a+6=0\),
\(\therefore\) 解得\(a=\sqrt{3}\),或\(2\sqrt{3}\).
故选:\(C\). -
答案 \(C\)
解析 由余弦定理可得, \(\cos C=\dfrac{a^2+b^2-c^2}{2 a b}=\dfrac{16+9-13}{2 \times 4 \times 3}=\dfrac{1}{2}\),
因为\(C\)为三角形的内角,故 \(C=\dfrac{1}{3} \pi\),
故选:\(C\). -
答案 \(AC\)
解析 由余弦定理知, \(\cos B=\dfrac{a^2+c^2-b^2}{2 a c}=\dfrac{a^2+c^2-\left(\dfrac{a+c}{\sqrt{3}}\right)^2}{2 a c}\),
\(∵B=\dfrac{\pi}{3}\),\(a+c=\sqrt{3} b\),
\(\therefore \dfrac{1}{2}=\dfrac{a^2+c^2-\left(\dfrac{a+c}{\sqrt{3}}\right)^2}{2 a c}\),化简得\(2a^2-5ac+2c^2=0\),
解得 \(a=\dfrac{1}{2} c\)或\(a=2c\),
\(\therefore \dfrac{a}{c}=\dfrac{1}{2}\)或\(2\).
故选:\(AC\). -
答案 \(120^∘\)
解析 因为在 中,设\(A\),\(B\),\(C\)和它们的对边\(a\),\(b\),\(c\),若\(a^2=b^2+c^2+bc\),
由余弦定理可知 \(\cos A=-\dfrac{1}{2}\),所以\(A=120^∘\). -
答案 \(10\)
解析 因为\(a=2\),\(b=4\), \(\cos C=\dfrac{1}{4}\),
由余弦定理得: \(c^2=a^2+b^2-2 a b \cos C=4+16-2 \times 2 \times 4 \times \dfrac{1}{4}=16\),
所以\(c=4\).
所以\(△ABC\)的周长为\(a+b+c=2+4+4=10\). -
答案 \(\dfrac{\pi}{3}\)
解析 \(a c \cos B=a^2-b^2+\dfrac{1}{2} b c\),
直接利用余弦定理 \(\cos B=\dfrac{a^2+c^2-b^2}{2 a c}\),
转换为\(b^2+c^2-a^2=bc\),整理得 \(\cos A=\dfrac{b^2+c^2-a^2}{2 b c}=\dfrac{1}{2}\),
由于\(0<A<π\),所以 \(A=\dfrac{\pi}{3}\). -
答案 \(\dfrac{2 \sqrt{7}}{7}\)
解析 \(△ABC\)中,角\(A\),\(B\),\(C\)和它们的对边\(a\),\(b\),\(c\),\(C=\dfrac{\pi}{3}\), \(b^2-c^2=\dfrac{1}{2} a^2\),
由余弦定理可得\(c^2=a^2+b^2-2ab⋅\cos C=a^2+b^2-ab\),
即\(b^2-c^2=ab-a^2\),
\(\therefore \dfrac{a^2}{2}=a b-a^2\), \(b=\dfrac{3 a}{2}\),
再把 \(b=\dfrac{3 a}{2}\)代入 \(b^2-c^2=\dfrac{1}{2} a^2\),可得 \(c=\dfrac{\sqrt{7}}{2} a\).
则 \(\cos A=\dfrac{b^2+c^2-a^2}{2 b c}=\dfrac{2 \sqrt{7}}{7}\). -
答案 \(\dfrac{\sqrt{6}}{2}\)
解析 如图所示,
\(△ABC\)中,\(c=2\),\(b=1\), \(\cos C=\dfrac{1}{4}\),
由余弦定理得,\(c^2=a^2+b^2-2ac\cos C\),即 \(4=a^2+1-2 a \times 1 \times \dfrac{1}{4}\),
整理得\(2a^2-a-6=0\),解得\(a=2\)或 \(a=-\dfrac{3}{2}\)(舍去);
所以 \(C D=\dfrac{1}{2} a=1\),
由余弦定理得, \(A D^2=12+12-2 \times 1 \times 1 \times \dfrac{1}{4}=\dfrac{3}{2}\),
解得 \(A D=\dfrac{\sqrt{6}}{2}\),
所以\(△ABC\)的中线AD的长为\(\dfrac{\sqrt{6}}{2}\).
故答案为:\(\dfrac{\sqrt{6}}{2}\). -
答案 (1) \(\dfrac{3}{5}\); (2) \(\sqrt{37}\).
解析 (1)在\(△ABC\)中,\(A B=\sqrt{10}\),\(BC=3\),\(AC=5\),
由余弦定理可得 \(\cos \angle A C B=\dfrac{A C^2+B C^2-A B^2}{2 A C \cdot B C}=\dfrac{25+9-10}{2 \times 5 \times 3}=\dfrac{4}{5}\),
\(\therefore \sin \angle A C B=\sqrt{1-\cos ^2 \angle A C B}=\dfrac{3}{5}\);
(2)结合(1)可知 \(\cos \angle A C D=\cos \left(135^{\circ}-\angle A C B\right)=\cos 135^{\circ} \cos \angle A C B+\sin 135^{\circ} \sin \angle A C B\)
\(=-\dfrac{\sqrt{2}}{2} \times \dfrac{4}{5}+\dfrac{\sqrt{2}}{2} \times \dfrac{3}{5}=-\dfrac{\sqrt{2}}{10}\),
\(\because C D=2 \sqrt{2}\),\(AC=5\),
\(\therefore\)在\(△ACD\)中,由余弦定理可得
\(A D=\sqrt{A C^2+C D^2-2 A C \cdot C D \cdot \cos \angle A C D}\)\(=\sqrt{5^2+(2 \sqrt{2})^2-2 \times 5 \times 2 \sqrt{2} \times\left(-\dfrac{\sqrt{2}}{10}\right)}=\sqrt{37}\).
【B组---提高题】
1.\(Δ ABC\)中三边上的高依次为\(\dfrac{1}{13}\), \(\dfrac{1}{5}\), \(\dfrac{1}{11}\),判定\(ΔABC\)的形状.
2.在\(△ABC\)中,\(∠C=90°\),\(M\)是\(BC\)边上一点,且满足 \(\overrightarrow{C M}=2 \overrightarrow{M B}\),若 \(\sin \angle B A M=\dfrac{1}{5}\),则\(\sin∠BAC=\)\(\underline{\quad \quad}\).
参考答案
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答案 钝角三角形
解析 设\(Δ ABC\)三边分别为\(a\),\(b\),\(c\), \(S_{\triangle A B C}=\dfrac{1}{2} a \cdot \dfrac{1}{13}=\dfrac{1}{2} b \cdot \dfrac{1}{11}=\dfrac{1}{2} c \cdot \dfrac{1}{5}\),
所以 \(\dfrac{a}{13}=\dfrac{b}{11}=\dfrac{c}{5}\),设\(a=13 k\),\(b=11 k\),\(c=5 k(k>0)\) .
因为\(11 k+5 k>13 k\),
故能构成三角形,取大角\(A\) ,
\(\cos A=\dfrac{b^2+c^2-a^2}{2 b c}=\dfrac{11^2+5^2-13^2}{2 \times 11 \times 5}<0\),
所以\(A\)为钝角,所以\(Δ ABC\)为钝角三角形. -
答案 \(\dfrac{\sqrt{15}}{5}\)
解析 记\(∠BAM=θ\),则 \(\sin \theta=\dfrac{1}{5}\),
设\(BC=3\),因\(\overrightarrow{C M}=2 \overrightarrow{M B}\),所以\(BM=1\),\(MC=2\),
设\(CA=t\),由\(∠C=90°\),得 \(A B=\sqrt{9+t^2}\), \(A M=\sqrt{4+t^2}\),
因 \(\sin \theta=\dfrac{1}{5}\),所以 \(\cos \theta=\dfrac{2 \sqrt{6}}{5}\),
因 \(B M^2=A B^2+A M^2-2 A B \cdot A M \cos \theta\),
即 \(1=9+t^2+4+t^2-2 \sqrt{9+t^2} \sqrt{4+t^2} \cdot \dfrac{2 \sqrt{6}}{5}\),
整理得:\(t^4-12t^2+36=0\),即\((t^2-6)^2=0\),所以\(t^2=6\),
所以 \(A B=\sqrt{9+t^2}=\sqrt{9+6}=\sqrt{15}\),
所以 \(\sin \angle B A C=\dfrac{3}{\sqrt{15}}=\dfrac{\sqrt{15}}{5}\).
故答案为: \(\dfrac{\sqrt{15}}{5}\).