Xét ΔABD có
\(cosDAB=\dfrac{AB^2+AD^2-BD^2}{2\cdot AB\cdot AD}\)
=>\(AB^2+AD^2-BD^2=2\cdot AB\cdot AD\cdot cosA\)
=>\(24^2+60^2-BD^2=2\cdot24\cdot60\cdot cos55\)
=>\(BD^2=4176-2880\cdot cos55\)
=>\(BD=\sqrt{4176-2880\cdot cos55}\simeq50,240\left(cm\right)\)