Do \(\widehat{M}=90^0\) nên tam giác MDE vuông tại M
Trong tam giác MDE:
\(tan\widehat{E}=\dfrac{MD}{ME}\Rightarrow MD=ME.tan\widehat{E}=20.tan60^0=20\sqrt{3}\left(m\right)\)
\(\Rightarrow DN=MD-MN=20\sqrt{3}-5\left(m\right)\)
Do CN song song ME \(\Rightarrow\widehat{DCN}=\widehat{E}=60^0\)
Trong tam giác vuông DCN:
\(sin\widehat{DCN}=\dfrac{DN}{DC}\Rightarrow DC=\dfrac{DN}{sin\widehat{DCN}}=\dfrac{20\sqrt{3}-5}{sin60^0}\approx34,23\left(m\right)\)