a, Áp dụng PTG: \(AC=\sqrt{BC^2-AB^2}=12\left(cm\right)\)
Ta có \(\sin\widehat{B}=\dfrac{AC}{BC}=\dfrac{12}{15}=\dfrac{4}{5}\approx\sin53^0\Leftrightarrow\widehat{B}\approx53^0\)
Vì \(\widehat{C}+\widehat{B}=90^0\Rightarrow\widehat{C}=90^0-\widehat{B}\approx90^0-53^0=37^0\)
\(b,AB\cdot\tan B+AC\cdot\tan C=AB\cdot\dfrac{AC}{AB}+AC\cdot\dfrac{AB}{AC}=AC+AB=21\left(cm\right)\)

