This (seemingly simple) problem is driving me nuts.
Find angle $\alpha$ shown in the following regular 18-gon.
It was easy to find the angle between pink diagonals ($60^\circ$). And I was able to solve the problem with some trigonometry (getting nice integer angle). However, all my attempts to solve the problem without use of trigonometry have failed. It looked like I was close to solution all the time (so many angles are equal to $60^\circ$ or $120^\circ$. I felt like I had to draw just one more line and the problem would break apart. I also tried with internal symmetries and rotations but eventually I had to give up.
Is there a way to solve this kind of problem without sines and cosines?
I don't have a direct solution. The best I have relates to a different problem, so I show the latter's solution, then show how the two problems correspond.
Lemma. Let $OAC$ be a triangle, and $X$ a point in it, and let angles $CAX=40^{\circ}, XAO=10^{\circ}, AOX=10^{\circ}, XOC=70^{\circ}$. What is $\angle OCX$?
Solution of lemma. $\angle AOC=80^{\circ}$ and $\angle CAO=50^{\circ}$ so $\angle OCA=50^{\circ}= \angle CAO$ so $\triangle ACO$ is isosceles on base $AC$, so $OA=OC$.
$\angle AOX = 10^{\circ} = \angle XAO$, so $\triangle OAX$ is isosceles on base $OA$, so $AX=OX$.
$\angle OXA = 180^{\circ} - \angle XAO - \angle AOX = 160^{\circ}$.
Erect an equilateral $\triangle OXE$ on base $OX$. Join $CE$. $\angle EOC = \angle XOC - \angle XOE = 70^{\circ} - 60^{\circ} = 10^{\circ} = \angle AOX$. Thus $\triangle$s $OAX, OCE$ are congruent (SAS, opposite sense) because $\angle EOC = \angle AOX, OA=OC$ and $OX=OE$. Thus $\angle OCE = \angle XAO = 10^{\circ}$. $CE=AX=OX=XE$. Thus $\triangle XCE$ is isosceles on base $XC$. $\angle XEC = 360^{\circ} - \angle CEO - \angle OEX = 360^{\circ} - 160^{\circ} - 60^{\circ} = 140^{\circ}$. Thus $\angle ECX = (180^{\circ}-140^{\circ})/2 = 20^{\circ}$. Thus $\angle OCX = \angle ECX + \angle OCE = 20^{\circ}+10^{\circ} = 30^{\circ}$, which solves the lemma.
Solution of the original problem. Let $P_0\dots P_{17}$ be a regular 18-gon. Let $P_1P_{13}$ cross $P_{5}P_{15}$ at $X$. What is $\alpha=\angle P_0XP_1$?
Let the centre of the 18-gon be $O$. $\triangle P_0P_{1}O$ is isosceles on base $P_{0}P_1$, and $\angle P_0OP_1=20^\circ$, so $\angle OP_1P_0=80^\circ$.
$\triangle P_{13}P_1O$ is isosceles on base $P_{13}P_1$, and $\angle P_{13}OP_1=120^\circ$, so $\angle OP_1P_{13}=30^\circ=\angle OP_1X$ as $X$ is on $P_1P_{13}$.
$\triangle P_{15}P_5O$ is isosceles on base $P_{15}P_5$, and $\angle P_{15}OP_5=160^\circ$, so $\angle P_5P_{15}O=10^\circ=\angle XP_{15}O$ as $X$ is on $P_5P_{15}$.
$\triangle P_{15}P_1O$ is isosceles on base $P_{15}P_1$, and $\angle P_{15}OP_1=80^\circ$, so $\angle P_1P_{15}O=50^\circ$. Thus this $\triangle$ has the same angles $80^\circ, 50^\circ, 50^\circ$ as the lemma's $\triangle OAC$, so they are similar. Moreover, the angles $OP_1X=30^\circ$ and $XP_{15}O=10^\circ$ correspond to those in the lemma, so the $X$s correspond. Therefore, by the lemma, $OX=XP_{15}$.
$OP_0=OP_{15}$ and $\angle P_{15}OP_0=60^\circ$ so $\triangle OP_{15}P_0$ is equilateral. Thus $\triangle$s $OXP_0$ and $P_{15}XP_0$ are congruent in opposite senses (SSS) because $OX=XP_{15}$, $OP_0=P_{15}P_0$ and $P_0X$ is common. Thus $\angle OP_0X=\angle XP_0P_{15}$ so $\angle OP_0X=30^\circ=\angle OP_1X$, so quadrilateral $OXP_0P_1$ is cyclic, so $\alpha=\angle P_0XP_1=\angle P_0OP_1=20^\circ$, which solves the problem.