Finding an unknown angle Geometry: Auxiliary Lines 
As shown in the figure: 


This is not the desired method, but it does show that $50^\circ$ is the result.  
Call the fourth point D, and suppose A is at $(0,0)$ and C is at $(1,0)$.  Then B is at $(x_B,y_B)=\left(\frac{\tan(70^\circ)}{\tan(58^\circ)+\tan(70^\circ)},\frac{\tan(58^\circ)\tan(70^\circ)}{\tan(58^\circ)+\tan(70^\circ)}\right) \approx (0.631921861,1.011286374)$ while D is at $(x_D,y_D)=\left(\frac{\tan(62^\circ)}{\tan(12^\circ)+\tan(62^\circ)},\frac{\tan(12^\circ)\tan(62^\circ)}{\tan(12^\circ)+\tan(62^\circ)}\right) \approx (0.898457801,0.190973101)$.
Then the length of AD is $a=\sqrt{x_D^2+y_D^2} \approx 0.918529883$ and of BD is $b = \sqrt{(x_D-x_B)^2+(y_D-y_C)^2} \approx 0.862528419$.  Using the sine rule, the angle ABD is $$\sin^{-1}\left(\frac{a}{b}\sin(46^\circ)\right) = 50^\circ$$ at least to the precision of my calculations.
For what it is worth, an accurate diagram looks like 

