\(a,AC=\sqrt{BC^2-AB^2}=16\left(cm\right)\left(pytago\right)\)
Áp dụng HTL: \(AH\cdot BC=AB\cdot AC\Leftrightarrow AH=\dfrac{192}{20}=9,6\left(cm\right)\)
\(\sin\widehat{B}=\dfrac{AC}{BC}=\dfrac{16}{20}=\dfrac{4}{5}\approx\sin53^07'\Leftrightarrow\widehat{B}\approx53^07'\)