Do \(AB = \dfrac{1}{3}BC\) nên \(\sin C = \dfrac{{AB}}{{BC}} = \dfrac{1}{3}.\) Từ đó
\(\eqalign{
& \cos C = \sqrt {1 - \dfrac{1}{9}} = \dfrac{{2\sqrt 2 }}{ 3}, \cr
& tgC = \dfrac{{\sin C}}{{\cos C}} = \dfrac{1}{ {2\sqrt 2 }} = \dfrac{{\sqrt 2 }}{4}, \cr
& \cot gC = \dfrac{4}{ {\sqrt 2 }} = 2\sqrt {2}. \cr} \)
