$from \: the \: figure \\ AB = 300 \: ft \: \: \: angle \: C = 60 \:( theta)\: degress \\ in \: triangle \: ABC \: using \: tan \: theta\\ \tan(60) = \frac{opposite \: side}{adjacent \: side} \\ \tan(60) = \frac{300}{BC} \\ \sqrt{3} = \frac{300}{BC} \\ BC = \frac{300}{ \sqrt{3} } \\ BC = 100 \sqrt{3} ft \\ then \: in \: triangle \: ABD \: using \: tan \: theta \\ angle \: D = 30 \: degrees \: (theta) \\ \tan(30 ) =\frac{opposite \: side}{adjacent \: side} \\ \tan(30) = \frac{300}{BD} \\ \frac{1}{ \sqrt{ 3} } = \frac{300}{BD} \\ BD = 300 \sqrt{3} \\ CD = BD - BC \\ = 300 \sqrt{3 } - 100 \sqrt{3} \\ = (300 - 100) \sqrt{3} \\ = 200 \sqrt{3 } \\ = 200 \times 1.732 \\ = 346.4 \: ft$

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