Câu b)$\widehat{BMC}=120^0 \Rightarrow \widehat{BMA}=60^0 \Rightarrow \widehat{ACE}=\widehat{ABM}=30^0$
$\triangle ACE$ vuông ở $A$
$\Rightarrow \frac{EA}{EC}=sin30^0=\frac{1}2$
$\Rightarrow \frac{S_{EAD}}{S_{EBC}}=(\frac{EA}{EC})^2$
$\Rightarrow S_{EBC}=40 cm^2$