# Finding the length of a side in an irregular quadrilateral, given three angle measures and two other sides I'm struggling with finding the length $$x$$. It is obvious that $$\angle DCB = 150^\circ$$.

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

## 2 Answers

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 $$x=9$$.

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

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