<s>[INST] Find angle UFO in the picture attached I sent this problem to Presh Talwalkar who suggested me to send it to this site.
I tried many things but was not able to find the correct solution.


*

*I made various segments trying to get an equilateral triangle similar to the Russian triangle problem, but no success. 

*I also tried to flip the triangle UFO over side NO but again no success.  

*I tried to find like triangles, but not enough. 
Could you please give me a hint?
Thanks,
R. de Souza

 [/INST] Without loss of generality, let $OE=1$.  By the Law of Sines on the triangle $ONE$, $ON=\dfrac{\sin(40^\circ)}{\sin(60^\circ)}$.  Thus, using the Law of Sines on the triangle $ONF$, we get
$$NF=ON\,\left(\frac{\sin(20^\circ)}{\sin(100^\circ)}\right)=ON\,\left(\frac{\sin(20^\circ)}{\sin(80^\circ)}\right)=\frac{\sin(20^\circ)\,\sin(40^\circ)}{\sin(60^\circ)\,\sin(80^\circ)}\,.$$
Furthermore, the Law of Sines on the triangle $OUE$ gives
$$OU=\frac{\sin(35^\circ)}{\sin(65^\circ)}\,.$$
We also have $NE=\dfrac{\sin(80^\circ)}{\sin(60^\circ)}$ (applying the Law of Sines on the triangle $ONE$), which gives
$$NU=NE\,\left(\frac{\sin(5^\circ)}{\sin(115^\circ)}\right)=NE\,\left(\frac{\sin(5^\circ)}{\sin(65^\circ)}\right)=\frac{\sin(80^\circ)\,\sin(5^\circ)}{\sin(60^\circ)\,\sin(65^\circ)}\,,$$
using the Law of Sines on the triangle $UNE$.
Thus,
$$\frac{NU}{NF}=\frac{\sin^2(80^\circ)\,\sin(5^\circ)}{\sin(20^\circ)\,\sin(40^\circ)\,\sin(65^\circ)}\,.\tag{*}$$
Note that
$$\sin(65^\circ)\,\sin(25^\circ)=\frac{1}{2}\,\big(\cos(40^\circ)-\cos(90^\circ)\big)=\frac{\cos(40^\circ)}{2}\,,$$
where we use the identity $\sin(x)\,\sin(y)=\dfrac{1}{2}\,\big(\cos(x-y)-\cos(x+y)\big)$.
Thus, (*) becomes
$$\frac{NU}{NF}=\frac{2\,\sin^2(80^\circ)\,\sin(5^\circ)\,\sin(25^\circ)}{\sin(20^\circ)\,\sin(40^\circ)\,\cos(40^\circ)}\,.$$
From the identity $\sin(2x)=2\,\sin(x)\,\cos(x)$, we get
$$\frac{NU}{NF}=\frac{4\,\sin(80^\circ)\,\sin(5^\circ)\,\sin(25^\circ)}{\sin(20^\circ)}=\frac{4\,\cos(10^\circ)\,\sin(5^\circ)\,\sin(25^\circ)}{\sin(20^\circ)}\,.$$
That is,
$$\frac{NU}{NF}=\frac{4\,\cos(10^\circ)\,\cos(5^\circ)\,\sin(5^\circ)\,\sin(25^\circ)}{\sin(20^\circ)\,\cos(5^\circ)}\,.$$
As $\sin(4x)=2\,\sin(2x)\,\cos(2x)=4\,\sin(x)\,\cos(x)\,\cos(2x)$, we get
$$\frac{NU}{NF}=\frac{\sin(25^\circ)}{\cos(5^\circ)}=\frac{\sin(25^\circ)}{\sin(95^\circ)}\,.$$
Ergo, if $\theta:=\angle UFN$, then we have from the Law of Sines on the triangle $UNF$ that
$$\frac{\sin(\theta)}{\sin(120^\circ-\theta)}=\frac{NU}{NF}=\frac{\sin(25^\circ)}{\sin(120^\circ-25^\circ)}\,.$$
It follows immediately from the identity $\sin(x)\,\sin(y)=\dfrac{1}{2}\,\big(\cos(x-y)-\cos(x+y)\big)$ that
$$\cos(120^\circ+25^\circ-\theta)=\cos(120^\circ-25^\circ+\theta)\,.$$
That is, 
$$25^\circ-\theta=n\cdot 180^\circ$$ for some integer $n$.  As $0^\circ<\theta<100^\circ$ (because $\angle OFN=100^\circ$), we have $n=0$, whence $\theta=25^\circ$.  That is,
$$\angle UFO=180^\circ-25^\circ-80^\circ=75^\circ\,.$$
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