I know that this question has been posted already, but I saw that is possible to draw the triangle of the question inside of the russian triangle. Can someone help me to finish the question using this?
I find a lot of angles and I got the equation $∠UFO+∠UFN=100°$, but I couldn't find the value of $∠UFN$
Let $OF\cap UE=\{D\}$ and $\measuredangle UFO=x$.
Thus, $\measuredangle DUF=95^{\circ}$ and since $$\frac{OD}{UD}\cdot\frac{UD}{FD}\cdot\frac{FD}{ED}\cdot\frac{ED}{OD}=1,$$ by law of sines we obtain: $$\frac{\sin65^{\circ}}{\sin20^{\circ}}\cdot\frac{\sin{x}}{\sin(95^{\circ}-x)}\cdot \frac{\sin5^{\circ}}{\sin80^{\circ}}\cdot \frac{\sin60^{\circ}}{\sin35^{\circ}}=1 $$ or $$\frac{\sin65^{\circ}\sin5^{\circ}\sin60^{\circ}}{\sin20^{\circ}\sin80^{\circ}\sin35^{\circ}}=\sin95^{\circ}\cot{x}-\cos95^{\circ}$$ or $$\frac{\sin65^{\circ}\sin5^{\circ}\sin60^{\circ}}{\sin20^{\circ}\sin80^{\circ}\sin35^{\circ}}-\sin5^{\circ}=\cos5^{\circ}\cot{x}$$ or $$\tan5^{\circ}\left(\frac{\sin65^{\circ}\sin60^{\circ}}{\sin20^{\circ}\sin80^{\circ}\sin35^{\circ}}-1\right)=\cot{x}.$$ We need to prove that: $$\tan5^{\circ}\left(\frac{\sin65^{\circ}\sin60^{\circ}}{\sin20^{\circ}\sin80^{\circ}\sin35^{\circ}}-1\right)=\tan15^{\circ}$$ or $$\frac{\sin65^{\circ}\sin60^{\circ}}{\sin20^{\circ}\sin80^{\circ}\sin35^{\circ}}-1=\frac{3-\tan^25^{\circ}}{1-3\tan^25^{\circ}}$$ or $$\frac{\sin65^{\circ}\sin60^{\circ}}{\sin20^{\circ}\sin80^{\circ}\sin35^{\circ}}=\frac{4-4\tan^25^{\circ}}{1-3\tan^25^{\circ}}$$ or $$\frac{\sin65^{\circ}\sin60^{\circ}}{\sin20^{\circ}\sin80^{\circ}\sin35^{\circ}}=\frac{8\cos10^{\circ}}{1+\cos10^{\circ}-3+3\cos10^{\circ}}$$ or $$\frac{\sin65^{\circ}\sin60^{\circ}}{\sin20^{\circ}\sin80^{\circ}\sin35^{\circ}}=\frac{2\cos10^{\circ}}{\cos10^{\circ}-\cos60^{\circ}}$$ or $$\frac{\cos25^{\circ}\sin60^{\circ}}{\sin20^{\circ}\cos10^{\circ}\sin35^{\circ}}=\frac{\cos10^{\circ}}{\sin25^{\circ}\sin35^{\circ}}$$ or $$\sin50^{\circ}\sin60^{\circ}=\sin20^{\circ}(1+\cos20^{\circ})$$ or $$\sin60^{\circ}\cos40^{\circ}-\cos60^{\circ}\sin40^{\circ}=\sin20^{\circ},$$ which is obvious.