In triangle ABD, [tex]\angle D[/tex] = 30°
Applying trigonometry,
[tex] \boxed {\mathrm{ \tan(30) = \dfrac{ AB}{BD} }}[/tex]
- [tex] \dfrac{1}{ \sqrt{3} } = \dfrac{100}{BD } [/tex]
- [tex]BD = 100 \sqrt{3} [/tex]
Now, In triangle ABC, [tex]\angle C[/tex] = 60°
Applying trigonometry,
[tex] \boxed{ \mathrm{ \tan(60) = \frac{AB}{BC} }}[/tex]
- [tex] \sqrt{3} = \dfrac{100}{BC} [/tex]
- [tex]BC = \dfrac{100}{ \sqrt{3} } [/tex]
Measure of CD = BD - BC
- [tex]CD = 100 \sqrt{3} - \dfrac{100}{ \sqrt{3} } [/tex]
- [tex]CD = \dfrac{300 - 100}{ \sqrt{3} } [/tex]
- [tex]CD = \dfrac{200}{ \sqrt{3} } [/tex]
- [tex]CD = 115.47 \: \: ft[/tex]
- [tex] \boxed {\mathrm{CD \approx115.5 \: \: ft}}[/tex]
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[tex]\mathrm{ ☠ \: TeeNForeveR \:☠ }[/tex]