Find angle $\alpha$ in triangle These is our teacher question. It's geometry question. Is it possible to help me to solve question. we need angle $\alpha$ in the triangle.   
Information:
$\angle FBG = 50, \angle GBC = 20 , \angle FCB = 50, \angle FCG=20$ 

Thanks.
 A: Let $AB=AC=1.$ Since $\angle BAC=40$ (degrees) we have $BC=2\sin 20.$ Note that  $BGC$ is a right angle.By the Sine Law for triangle $BGC,$ we have $  CG/\sin 20=CG/ \sin CBG=BC/\sin BGC=BC/\sin 90=BC ,$ so $$(i)\quad CG=BC\sin 20=2 \sin^2 20.$$  Therefore $$(ii)\quad AG=1-CG=1-2 \sin^2 20=\cos 40.$$ We have $\angle BFC=60.$ By the Sine Law for triangle $BFC$, we have $BF/\sin 50=BF/\sin BCF=BC/\sin BFC=BC/\sin 60,$ so $BF=BC \sin 50/\sin 60.$ Therefore $$(iii) \quad AF=1-BF=1-BC\sin 50/\sin 60=1-2\sin 20 \sin 50/\sin 60.$$ We have $\angle FAG=\angle BAC=40.$ By the Cosine Law for triangle $FAG$ and by (ii) and (iii),therefore $$(iv)\quad FG^2=AF^2+AG^2-2AF\cdot AG\cos FAG=AF^2+AG^2-2AF\cdot AG\cos 40.$$ By the Sine Law for triangle $FGC$ we have $\sin \alpha/CG=\sin FCG/FG=\sin 20/FG$ so from (i) we have $$(v)\quad \sin \alpha=CG\sin 20/FG=2  \sin^320/FG.$$ Now in (v) the value FG is obtained from (iv), and in (iv) the values AF, AG are obtained from (ii) and (iii).
A: Suppose that $|BC| = 1$. Then, triangle $BGC$ is right with hypotenuse $1$, so $|BG| = \sin 70^\circ$ and $CG = \sin 20^\circ$. Use the law of sines on triangle $BFC$ to see that $|BF| = \frac{\sin 50^\circ}{\sin 60^\circ}$. The law of cosines then implies that $|FG| = \sqrt{|BF|^2 + |BG|^2 - 2|BF||BG|\cos 50^\circ}$.
Finally, the law of sines gives $\alpha = \arcsin\left(\frac{(\sin 20^\circ)^2}{|FG|}\right) \approx 8.709^\circ$
A: 
I draw the triangle hopeful to get a whole angle, but no.
