Find the highlighted angle in the regular enneagon inscribed below

Question

For reference: Let AB, BC be two adjacent sides of a regular 9-sided polygon inscribed in a circle with center O, as shown in the figure below. Let M be the midpoint of AB and N be the midpoint of the radius OT perpendicular to BC. Determine the measure, in degrees, of the angle OMN = alpha.(A:$30^o$)

OBs: The question asks not to use the law of sines or cosines

I found the possible angles but it lacks some relation to finish

$\triangle AOB_{isosceles}$

$\angle AOB = 40^o\\ \therefore \angle OAB=\angle OBA = \frac{180^o-40^0}{2} = 70^o\\ \angle MOB = \angle BOK = 20^o $

Answer 1

As shown, $\triangle OAT$ is equilateral and $AN$ is perpendicular to $OT$. $\angle OAN = 30^\circ$.

Since $AN$ is parallel to $BK$, $\angle NAB=40^\circ$. Also notice that $\angle NOM=40^\circ$, so we have $\angle NAM = \angle NOM$ so $A,M,N,O$ are co-cyclic. Therefore $\angle OMN=\angle OAN=30^\circ$

Answer 2

An alternative approach would be to use the law of sines in $\triangle OMN$. You know $ON=0.5r$ and $OM=r\cos20^\circ$. If $\alpha=\angle OMN$, then $\angle ONM=140^\circ-\alpha$. Therefore: $$\frac{\sin\alpha}{0.5r}=\frac{\sin(140^\circ-\alpha)}{r\cos20^\circ}$$ It's easy to get from here to $$\tan\alpha=\frac{0.5\sin140^\circ}{0.5\cos140^\circ+\cos20^\circ}$$ I've used a calculator to get $\alpha=30^\circ$, so I assume that there is a way to simplify the last fraction.