A 30˚ angle inside an Equilateral Triangle

Draw $\overline{BP}$ and $\overline{CP}$ and note that $\angle PBC=30^\circ$ and $\angle PCB=30^\circ$ . Note that $\angle PDE=\angle PBD+\angle BPD=\angle DPE+\angle BPD=\angle BPE$ . So $\triangle PBE\sim\triangle DCP$ which implies ${s\over 12}={(x+7)\over s}$ or $s^2=12(x+7)$ .

Note that $\angle PBC=30^\circ$ , so $\cos 30^\circ=\frac{(5+7+x)/2}{s}=\frac{1}{2}\sqrt{3}$ . Hence $3s^2=(5+7+x)^2$ .

Combining these two equations gives $(5+7+x)^2=36(x+7)$ . Solving this equation for x and discarding the negative solution, gives $x=18$ .