Giải thích các bước giải:

a.Ta có:

$\cot B=\dfrac58$

$\to \dfrac{BA}{AC}=\dfrac58$

$\to \dfrac5{AC}=\dfrac58$

$\to AC=8$

$\to BC=\sqrt{AB^2+AC^2}=\sqrt{89}$

 b.Ta có:

$\sin B=\dfrac{AC}{BC}=\dfrac{8}{\sqrt{89}}$

$\cos B=\dfrac{BA}{BC}=\dfrac5{\sqrt{89}}$

$\tan B=\dfrac{AC}{AB}=\dfrac85$

$\cot B=\dfrac58$

Vì $\hat B+\hat C=90^o$

$\to \sin C=\cos B=\dfrac5{\sqrt{89}}$

      $\cos C=\sin B=\dfrac8{\sqrt{89}}$

     $\tan C=\cot B=\dfrac58$

     $\cot C=\tan B=\dfrac85$