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a, Tìm được sinα = 24 5 , tanα = 24 , cotα = 1 24
b, cosα = 5 3 , tanα = 2 5 , cotα = 5 2
c, sinα = ± 2 5 , cosα = ± 1 5 , cotα = 1 2
d, sinα = ± 1 10 , cosα = ± 3 10 , tanα = 1 3
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\cos\alpha=\dfrac{1}{2}\)
\(\tan\alpha=\sqrt{3}\)
\(\cot\alpha=\dfrac{\sqrt{3}}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: tan B=3/4
=>AC/AB=3/4
=>AC=9cm
BC=căn 9^2+12^2=15cm
b: sin B=căn 3/2
=>AC/AB=căn 3/2
=>AC=căn 3
BC=căn AB^2+AC^2=2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(tan\alpha=3\Rightarrow cot\alpha=\dfrac{1}{3}\left(tan\alpha.cot\alpha=1\right)\)
\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Rightarrow cos^2\alpha=\dfrac{1}{1+tan^2\alpha}=\dfrac{1}{1+9}=\dfrac{1}{10}\)
\(\Rightarrow cos^{ }\alpha=\dfrac{\sqrt[]{10}}{10}\)
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}\Rightarrow sin\alpha=tan\alpha.cos\alpha=\dfrac{3\sqrt[]{10}}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,HC=\dfrac{AH^2}{BH}=\dfrac{256}{9}\\ \Rightarrow AB=\sqrt{BH\cdot BC}=\sqrt{\left(\dfrac{256}{9}+9\right)9}=\sqrt{337}\\ 2,BC=\sqrt{AB^2+AC^2}=10\left(cm\right)\\ \Rightarrow BH=\dfrac{AB^2}{BC}=6,4\left(cm\right)\\ 3,AC=\sqrt{BC^2-AB^2}=9\\ \Rightarrow CH=\dfrac{AC^2}{BC}=5,4\\ 4,AC=\sqrt{BC\cdot CH}=\sqrt{9\left(6+9\right)}=3\sqrt{15}\\ 5,AC=\sqrt{BC^2-AB^2}=4\sqrt{7}\left(cm\right)\\ \Rightarrow AH=\dfrac{AB\cdot AC}{BC}=3\sqrt{7}\left(cm\right)\\ 6,AC=\sqrt{BC\cdot CH}=\sqrt{12\left(12+8\right)}=4\sqrt{15}\left(cm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi `3x,2x(cm) (x \in NN^(**))` là độ dài của `AB,AC`.
Áp dụng hệ thức lượng:`
`1/(AH^2)=1/(AB^2)+1/(AC^2)`
`<=>1/(36^2)=1/(9x^2)+1/(4x^2)`
`<=>` \(\left[{}\begin{matrix}x=0\left(L\right)\\x=6\sqrt{13}\left(TM\right)\end{matrix}\right.\)
`=> AB=18\sqrt13 (cm) ; AC = 12\sqrt13 (cm)`
Áp dụng định lý Pitago:
`BC^2=AB^2+AC^2`
`<=>BC=7(cm)`
Áp dụng hệ thức lượng:
`AB^2=BH.BC`
`=> BH=54(cm)`
`=> CH=BC-BH=24 (cm)`
Vậy `BH=54cm ; CH=24cm`.
Ta có: \(\dfrac{AB}{AC}=\dfrac{3}{2}\)
nên \(AB=\dfrac{3}{2}AC\)
Xét ΔABC vuông tại A có
\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\)
\(\Leftrightarrow\dfrac{1}{36^2}=\dfrac{1}{\left(\dfrac{3}{2}AC\right)^2}+\dfrac{1}{AC^2}\)
\(\Leftrightarrow\dfrac{13}{9}\cdot\dfrac{1}{AC^2}=\dfrac{1}{1296}\)
\(\Leftrightarrow\dfrac{1}{AC^2}=\dfrac{1}{1872}\)
\(\Leftrightarrow AC^2=1872\)
hay \(AC=12\sqrt{13}\)(cm)
Ta có: \(AB=\dfrac{3}{2}\cdot AC\)
nên \(AB=\dfrac{3}{2}\cdot12\sqrt{13}=18\sqrt{13}\left(cm\right)\)
Áp dụng định lí Pytago vào ΔABH vuông tại H, ta được:
\(AB^2=AH^2+BH^2\)
\(\Leftrightarrow BH^2=\left(18\sqrt{13}\right)^2-36^2=2916\)
hay BH=54(cm)
Xét ΔABC vuông tại A có
\(AH^2=HB\cdot HC\)
\(\Leftrightarrow HC=\dfrac{AH^2}{BH}=\dfrac{36^2}{54}=24\left(cm\right)\)
Vậy: BH=54cm; CH=24cm
![](https://rs.olm.vn/images/avt/0.png?1311)
Có sin2a + cos2a = 1
Mà cos a = \(\dfrac{3}{4}\)
=> sin2a + (\(\dfrac{3}{4}\))2 = 1
=> sin2a + \(\dfrac{3^2}{4^2}\) = 1
=> sin2a + \(\dfrac{9}{16}\)= 1
=> sin2a = \(\dfrac{7}{16}\)
=> sin a = \(\dfrac{\sqrt{7}}{4}\)
Có tan a = \(\dfrac{\text{sin a}}{\text{cos a}}\)
Mà \(\left\{{}\begin{matrix}\text{cos a = }\dfrac{3}{4}\\\text{sin a = }\dfrac{\sqrt{7}}{4}\end{matrix}\right.\)
=> tan a = \(\dfrac{\dfrac{\sqrt{7}}{4}}{\dfrac{3}{4}}\) = \(\dfrac{\sqrt{7}}{4}\): \(\dfrac{3}{4}\) = \(\dfrac{\sqrt{7}}{4}\).\(\dfrac{4}{3}\) =\(\dfrac{\sqrt{7}}{3}\)