\(BC=\dfrac{5}{3}\cdot AB=\dfrac{100}{3}\left(cm\right)\)
\(CH=BC-BH=\dfrac{100}{3}-12=\dfrac{64}{3}\left(cm\right)\)
\(AH=\sqrt{20^2-12^2}=16\left(cm\right)\)
\(AC=\sqrt{16^2+\left(\dfrac{64}{3}\right)^2}=\dfrac{80}{3}\left(cm\right)\)
Xét ΔABC có \(AB^2+AC^2=BC^2\)
nên ΔABC vuong tại A
=>\(\widehat{BAC}=90^0\)