c) Ta có: $\widehat{CAH}+\widehat{EAH}=90^{0}$ (hai góc phụ nhau)
$\Rightarrow \widehat{EAH}=90^{0}-\widehat{CAH}=90^{0}-30^{0}=60^{0}$
Xét $\Delta AHE$ có: $\widehat{AHE}=90^{0}$
$\Rightarrow \widehat{HAE}+\widehat{HEA}=90^{0}$
$\Rightarrow \widehat{HEA}=90^{0}-\widehat{HAE}=90^{0}-60^{0}=30^{0}$
Vậy $\widehat{AEC}=30^{0}$