\(\sin\widehat{A}=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)

\(\cot\widehat{A}=\dfrac{5}{13}:\dfrac{12}{13}=\dfrac{5}{12}\)

\(\tan\widehat{B}=\dfrac{5}{12}\)

Có:

\(cosC=sinB=\dfrac{5}{13}\)

\(cosB=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\\ \Rightarrow sinC=\dfrac{12}{13}\)

\(tgC=\dfrac{sinC}{cosC}=\dfrac{\dfrac{12}{13}}{\dfrac{5}{13}}=\dfrac{12}{5}\)

\(\Rightarrow cotgC=\dfrac{5}{12}\)

Lời giải:

Ta có:

$\frac{5}{13}=\cos A=\frac{AC}{AB}$

$\Rightarrow AB=\frac{13}{5}AC$

Áp dụng định lý Pitago:

$AC^2+BC^2=AB^2$

$\Leftrightarrow AC^2+10^2=(\frac{13}{5}AC)^2$
$\Leftrightarrow 100=\frac{144}{25}AC^2$
$\Leftrightarrow AC^2=\frac{625}{36}$

$\Rightarrow AC=\frac{25}{6}$ (cm) 

Vậy......

Hình vẽ:

b: Áp dụng hệ thức lượng trong tam giác vuông vào ΔABC vuông tại A có AH là đường cao ứng với cạnh huyền BC, ta được:

\(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot BC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AB^2=21\\AC^2=28\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AB=\sqrt{21}\left(cm\right)\\AC=2\sqrt{7}\left(cm\right)\end{matrix}\right.\)

Xét ΔABC vuông tại A có 

\(\sin\widehat{B}=\cos\widehat{C}=\dfrac{AC}{BC}=\dfrac{2\sqrt{7}}{7}\)

\(\cos\widehat{B}=\sin\widehat{C}=\dfrac{AB}{BC}=\dfrac{\sqrt{21}}{7}\)

\(\tan\widehat{B}=\cot\widehat{C}=\dfrac{AC}{AB}=\dfrac{2\sqrt{7}}{\sqrt{21}}=\dfrac{2\sqrt{3}}{3}\)

\(\cot\widehat{B}=\tan\widehat{C}=\dfrac{AB}{AC}=\dfrac{\sqrt{21}}{2\sqrt{7}}=\dfrac{\sqrt{3}}{2}\)

cosB=0,8=4/5 => BA=4 , BC=5

Áp dụng định lý Pytago trong tam giác vuông ABC, có:

AC²=BC²-BA²

⁽⁼⁾ AC²=5²-4²=9

(=) AC=3

Ta có:

sinC=BA/BC=4/5

cosC=AC/BC=3/5

tanC=BA/AC=4/3

cotC=AC/BA=3/4

\(sin^2B+cos^2B=1\Leftrightarrow sin^2B-1-\left(0,8\right)^2=0.36.\Leftrightarrow sinB=0,6.\\\)

\(tanB=\frac{sinB}{cosB}=\frac{0,6}{0,8}=\frac{3}{4}\)

\(cotB=\frac{1}{tanB}=\frac{1}{\frac{3}{4}}=\frac{4}{3}.\)

\(sinC=cosB=0,8\)

\(cosC=sinB=0,6\)

\(tanC=cotB=\frac{4}{3}\)

\(cotC=tanB=\frac{3}{4}.\)

Xét ΔABC vuông tại A có

\(sinB=sin56\simeq0,83\)

\(cosB=cos56\simeq0,56\)

\(tanB=tan56\simeq1,48\)

\(cotB=cot56\simeq0,67\)

Xét ΔABC vuông tại A có

\(cosC=sinB\simeq0,83\)

\(sinC=cosB\simeq-0,56\)

\(cotC=tanB=tan56\simeq1,48\)

\(tanC=cotB\simeq0,67\)

\(\sin^2\widehat{A}+\cos^2\widehat{A}=1\Leftrightarrow\cos^2\widehat{A}=1-\left(\dfrac{3}{5}\right)^2=1-\dfrac{9}{25}=\dfrac{16}{25}\\ \Leftrightarrow\cos\widehat{A}=\dfrac{4}{5}\\ \tan\widehat{A}=\dfrac{\sin\widehat{A}}{\cos\widehat{A}}=\dfrac{3}{4}\\ \Rightarrow\cot\widehat{A}=\dfrac{1}{\tan\widehat{A}}=\dfrac{4}{3}\)