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

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

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

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:

\(a,cosC=\dfrac{5}{13}\\ Ta,có:cos^2C+sin^2C=1\\ \Rightarrow sinC=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\\ cosB+sinC=1\\ \Leftrightarrow cosB+\dfrac{12}{13}=1\\ \Rightarrow cosB=\dfrac{1}{13}\\ tanC=\dfrac{sinC}{cosC}=\dfrac{\dfrac{12}{13}}{\dfrac{5}{13}}=\dfrac{12}{5}\)

\(b,tanB=\dfrac{1}{5}\Rightarrow\dfrac{sinB}{cosB}=\dfrac{1}{5}\Rightarrow cosB=5sinB\\ E=\dfrac{sinB-3cosB}{2sinB+3cosB}=\dfrac{sinB-3.5.sinB}{2sinB+3.5.sinB}=\dfrac{-14sinB}{17sinB}=-\dfrac{14}{17}\)

Bạn ơi bạn làm được bài này chưa ạ. Chỉ mình với 

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

Ta có: \(\sin^2A+cos^2A=1\)

=>\(cos^2A=1-\left(\frac35\right)^2=1-\frac{9}{25}=\frac{16}{25}\)

=>\(cosA=\frac45\)

tan A=sin A:cosA

\(=\frac35:\frac45=\frac35\times\frac54=\frac34\)

\(\cot A=1:\frac34=\frac43\)

TA có: \(\sin^2A+cos^2A=1\)

=>\(cos^2A=1-\left(\frac35\right)^2=1-\frac{9}{25}=\frac{16}{25}=\left(\frac45\right)^2\)

=>\(cosA=\frac45\)

tan A=\(\frac{\sin A}{cosA}=\frac35:\frac45=\frac34\)

\(\cot A=\frac{1}{\tan A}=1:\frac34=\frac43\)