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I'm struggling with finding the length $x$. It is obvious that $\angle DCB = 150^\circ$.

  • $\begingroup$ Extend $AD$ and $BC$ to meet at $E$. $ABE$ is then an equilateral triangle. $\endgroup$ – FredH Mar 29 at 20:46

Extend $\overline{AD}$ and $\overline{BC}$ to form an equilateral triangle $\triangle ABE$. Then notice that $\triangle EDB$ is a $30$-$60$-$90$ right triangle whose $\sqrt{3}$ side is $5\sqrt{3}$. Therefore, the length of $\overline{DE}$ is $5$, the length of $\overline{BE}$ is $10$ and $x=9$.

  • $\begingroup$ Is it possible that you show it on a diagram? $\endgroup$ – Bobtrollsten Mar 29 at 20:51
  • $\begingroup$ @Bobtrollsten If you follow the directions given, then you will see for yourself how this works. $\endgroup$ – John Douma Mar 29 at 20:59

Consider $P\in AD$ such that $CP\parallel AB$. Then you have that $\angle DCP=30°$. Finally $$\tan(\angle DCP)=\tan(30°)=\frac{1}{\sqrt3}=\frac{x-4}{5\sqrt3}$$ Thus

$$5=x-4\iff \color{red}{x=9}$$


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