Xét \(\Delta AEB\) vuông tại E:
\(\widehat{BAE}+\widehat{ABE}=90^o\) (2 góc phụ nhau)
\(\Rightarrow\widehat{ABE}=90^o-30^o=60^o\)
Ta có: \(tan\widehat{ABE}=\dfrac{AE}{BE}\)
\(\Rightarrow BE=\dfrac{AE}{tan\widehat{ABE}}=\dfrac{35}{tan60^o}=\dfrac{35\sqrt{3}}{3}m\) (do \(AE=DC=35m\))
\(\Rightarrow BC=BE+EC=\dfrac{35\sqrt{3}}{3}+1,5=\dfrac{9+70\sqrt{3}}{6}m\)