\(\cot\widehat{C}=\dfrac{AC}{AB}=\dfrac{7}{24}\Rightarrow AB=\dfrac{14\cdot24}{7}=48\left(cm\right)\)
Áp dụng pytago:
\(BC=\sqrt{AB^2+AC^2}=50\left(cm\right)\)
\(\tan\widehat{C}=\dfrac{1}{\cot\widehat{C}}=\dfrac{24}{7}\\ \sin\widehat{C}=\dfrac{AB}{BC}=\dfrac{48}{50}=\dfrac{24}{25}\\ \cos\widehat{C}=\dfrac{AC}{BC}=\dfrac{14}{50}=\dfrac{7}{25}\)