\(\widehat{xBC}=\widehat{BCA}=20^o\left(So.le.trong\right)\)
\(\widehat{xBD}=\widehat{BDA}=30^o\left(So.le.trong\right)\)
\(tan\widehat{BCA}=\dfrac{AB}{CA}\Rightarrow CA=\dfrac{AB}{tan20^o}=\dfrac{75}{0,36}\approx208,3\left(m\right)\)
\(tan\widehat{BDA}=\dfrac{AB}{DA}\Rightarrow DA=\dfrac{AB}{tan30^o}=\dfrac{75}{0,58}\approx129,3\left(m\right)\)
\(CD=CA-DA=208,3-129,3=79\left(m\right)\)