\(\tan\widehat{ABC}=\dfrac{AC}{AB}=\dfrac{24}{15}=\dfrac{8}{5}\approx\tan58^0\Leftrightarrow\widehat{ABC}\approx58^0\)
Vì D là trung điểm AC nên \(AD=\dfrac{1}{2}AC=12\left(cm\right)\)
\(\tan\widehat{ABD}=\dfrac{AD}{AB}=\dfrac{12}{15}=\dfrac{4}{5}\approx\tan39^0\Leftrightarrow\widehat{ABD}\approx39^0\\ \Rightarrow\widehat{DBC}=\widehat{ABC}-\widehat{ABD}\approx58^0-39^0=19^0\)