cho tag = 5/3. Tính các tỉ số lượng giác còn lại
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\cos\alpha=\sqrt{1-\dfrac{1}{25}}=\dfrac{2\sqrt{6}}{5}\)
\(\tan\alpha=\dfrac{1}{5}:\dfrac{2\sqrt{6}}{5}=\dfrac{1}{2\sqrt{6}}=\dfrac{\sqrt{6}}{12}\)
\(\cot\alpha=1:\dfrac{1}{2\sqrt{6}}=2\sqrt{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\cos\alpha=0.8\)
\(\tan\alpha=\dfrac{3}{4}\)
\(\cot\alpha=\dfrac{4}{3}\)
\(sina=0,6\Rightarrow cosa=\sqrt{1-sin^2a}=\sqrt{1-0,6^2}=0,8\)
\(tana=\dfrac{sina}{cosa}=\dfrac{0,6}{0,8}=\dfrac{3}{4}\)
\(cota=\dfrac{1}{tana}=\dfrac{4}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\tan\alpha=\dfrac{1}{3}\Leftrightarrow1+\tan^2\alpha=\dfrac{1}{\cos^2\alpha}\\ \Leftrightarrow\cos\alpha=\dfrac{1}{\sqrt{1+\tan^2\alpha}}=\dfrac{1}{\sqrt{\dfrac{9}{8}}}=\dfrac{2\sqrt{2}}{3}\\ \sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-\dfrac{8}{9}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\\ \tan\alpha=\dfrac{1}{3}\Leftrightarrow\cot\alpha=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
sin a=12/13
cos^2a=1-(12/13)^2=25/169
=>cosa=5/13
tan a=12/13:5/13=12/5
cot a=1:12/5=5/12
sin b=căn 3/2
cos^2b=1-(căn 3/2)^2=1/4
=>cos b=1/2
tan b=căn 3/2:1/2=căn 3
cot b=1/căn 3
![](https://rs.olm.vn/images/avt/0.png?1311)
sinB = b/a; cosB = c/a; tgB = b/c; cotgB = c/b
sinC = c/a; cosC = b/a; tgC = c/b; cotgB = b/c
a) b = a.(b/a) = a.sinB = a.cosC
c = a. (c/a) = a.cosB = a.sinC
b) b = c. (b/c) = c.tgB = c.cotgC
c = b.(c/b) = b.cotgB = b.tgC
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\cot\widehat{C}=\dfrac{AC}{AB}=\dfrac{7}{24}\Rightarrow AB=\dfrac{14\cdot24}{7}=48\left(cm\right)\)
Áp dụng pytago:
\(BC=\sqrt{AB^2+AC^2}=50\left(cm\right)\)
\(\tan\widehat{C}=\dfrac{1}{\cot\widehat{C}}=\dfrac{24}{7}\\ \sin\widehat{C}=\dfrac{AB}{BC}=\dfrac{48}{50}=\dfrac{24}{25}\\ \cos\widehat{C}=\dfrac{AC}{BC}=\dfrac{14}{50}=\dfrac{7}{25}\)
Đề là: \(tan=\frac{5}{3}\)?
Ta có: Vì \(tan=\frac{5}{3}\) => \(cot=\frac{1}{tan}=\frac{1}{\frac{5}{3}}=\frac{3}{5}\)
Đặt 2 cạnh góc vuông còn lại là \(\hept{\begin{cases}5k\\3k\end{cases}\left(k>0\right)}\)
=> Cạnh huyền là: \(\sqrt{\left(5k\right)^2+\left(3k\right)^2}=\sqrt{34}k\)
=>\(sin=\frac{5k}{\sqrt{34}k}=\frac{5\sqrt{34}}{34}\) ; \(cos=\frac{3k}{\sqrt{34}k}=\frac{3\sqrt{34}}{34}\)
mới học k bt đúng hay sai, nếu sai thì thông cảm nhé
\(cota=\frac{1}{tana}\)
\(cota=\frac{1}{\frac{5}{3}}=\frac{3}{5}\)
\(1+tan^2a=\frac{1}{cos^2a}\)
\(1+\left(\frac{3}{5}\right)^2=\frac{1}{cos^2a}\)
\(1+\frac{9}{25}=\frac{1}{cos^2a}\)
\(\frac{34}{25}=\frac{1}{cos^2a}\)
\(cos^2a=\frac{25}{34}\)
\(cosa=\pm\sqrt{\frac{25}{34}}=\pm\frac{5\sqrt{34}}{34}\)
\(sin^2a+cos^2a=1\)
\(sin^2a+\frac{25}{34}=1\)
\(sin^2a=\frac{9}{34}\)
\(sina=\pm\sqrt{\frac{9}{34}}=\pm\frac{3\sqrt{34}}{34}\)