Theo Pitago, ta có:
\(AB^2+AC^2=BC^2\)
\(\Rightarrow AB^2=63\Rightarrow AB=\sqrt{63}\)
Các tỉ số lượng giác:
\(\sin C=\dfrac{AB}{BC}=\dfrac{\sqrt{63}}{12}=\dfrac{\sqrt{7}}{4}\)
\(\cos C=\dfrac{AC}{BC}=\dfrac{9}{12}=\dfrac{3}{4}\)
\(\text{tg }C=\dfrac{AB}{AC}=\dfrac{\sqrt{63}}{9}=\dfrac{\sqrt{7}}{3}\)
\(\text{cotg }C=\dfrac{AC}{AB}=\dfrac{9}{\sqrt{63}}=\dfrac{3}{\sqrt{7}}\)