Find $x$ $$\sin60 \sin30 \sin x=\sin50 \sin20 \sin(20-x)$$
$\frac{\sin(20^\circ-x)}{\sin x} = \cos 20^\circ-\sin20^\circ\cdot \cot x$, so we can easily find $x$. But you write Cheva theorem, probably, there exists a geometric solution.
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emonHR
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Lambert macuse
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Ceva's theorem is trivially satisfied as $(60^°+20^°)+(50^°+30^°)+(20^°-x+x)=180^°$. – Jon Nov 29 '19 at 13:47
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Is it related to https://math.stackexchange.com/questions/3469866/solve-sin12%c2%ba-sin24%c2%ba-sin84%c2%ba-x%c2%ba-sin30%c2%ba-sin30%c2%ba-sinx%c2%ba – lab bhattacharjee Dec 10 '19 at 13:34