A chord of length $6$ subtends an $80^\circ$ central angle in a circle. Can we calculate the distance from center to chord without trigonometry?

I know that the following can be answered easily using trigonometric ratios, but is there any way to go about it without relying on trigonometry? (The book from which the problem was taken doesn't cover trig before this point.)

$$O$$ is the centre of a circle, and $$PQ$$ and $$RS$$ are two equal length chords in it. Points $$M$$ and $$N$$ are the midpoints of the two chords $$PQ$$ and $$RS$$ respectively. The length of $$PQ$$ is given as $$6$$cm and the measure of the angle $$\angle POQ$$ (where $$O$$ is the vertex) is given as $$80^\circ$$ degrees. A diagram of this is in the snapshot I've attached. The question is, how do you calculate the length $$|OM|$$?

• What is the relevance of the RS chord? And what's the problem with trigonometry? – rafa11111 Dec 28 '18 at 13:46
• Well, the length of $OM$ (in centimeters) is $3\cot 40^\circ$. Computing this cotangent requires solving a degree 3 equation, because you need to trisect the angle of $120^\circ$. – egreg Dec 28 '18 at 13:50
• @egreg According to Wolfram, $3 \cot 40^\circ$ has the minimal polynomial $3x^6 -26x^4 + 33x^2 - 1$. – Connor Harris Dec 28 '18 at 15:34
• @ConnorHarris Yes, but this reduces to a degree 3 equation, doesn't it? – egreg Dec 28 '18 at 15:40
• And thanks goes to @Blue for that bit of editing which has considerably improved the presentation :) – Indula Dec 30 '18 at 10:25