Lời giải:
1. $BC=\sqrt{AB^2+AC^2}=\sqrt{3^2+4^2}=5$
$\sin B=\frac{AC}{BC}=\frac{4}{5}=\cos C$
$\cos B=\frac{AB}{BC}=\frac{3}{5}=\sin C$
$\tan B=\frac{AC}{AB}=\frac{4}{3}=\cot C$
$\cot B=\frac{AB}{AC}=\frac{3}{4}=\tan C$
2. $AB=\sqrt{BC^2-AC^2}=\sqrt{12^2-5^2}=\sqrt{119}$
$\sin B=\frac{AC}{BC}=\frac{5}{12}=\cos C$
$\cos B=\frac{AB}{BC}=\frac{\sqrt{119}}{12}=\sin C$
$\tan B=\frac{AC}{AB}=\frac{5}{\sqrt{119}}=\cot C$
$\cot B=\frac{AB}{AC}=\frac{\sqrt{119}}{5}=\tan C$
3.
$AC=\sqrt{BC^2-AB^2}=\sqrt{10^2-6^2}=8$
$\sin B=\frac{AC}{BC}=\frac{8}{10}=\frac{4}{5}=\cos C$
$\cos B=\frac{AB}{BC}=\frac{6}{10}=\frac{3}{5}=\sin C$
$\tan B=\frac{AC}{AB}=\frac{8}{6}=\frac{4}{3}=\cot C$
$\cot B=\frac{AB}{AC}=\frac{3}{4}=\tan C$
