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

 A: 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 

A: Let central point is $O$.
$\angle OBC = 180^\circ - (46^\circ+12^\circ) - (62^\circ+8^\circ) - X = 52^\circ-X$.
Using the sine formula, we have:
$$\frac{AO}{\sin 62^\circ} = \frac{CO}{\sin 12^\circ},$$
$$\frac{AO}{\sin X} = \frac{BO}{\sin 46^\circ},$$
$$\frac{CO}{\sin (52^\circ-X)} = \frac{BO}{\sin 8^\circ},$$
so,
$$\sin 12^\circ \cdot \sin 8^\circ \cdot \sin X = \sin 62^\circ \cdot \sin 46^\circ \cdot \sin (52^\circ-X).$$
(Here we can find that $X=50^\circ$. Uniqueness of solution is discussed in comments).

In fact we need to prove that
$$\sin 8^\circ \cdot \sin 12^\circ \cdot \sin 50^\circ \mathop{=}\limits^{???} \sin 2^\circ \cdot \sin 46^\circ \cdot \sin 62^\circ.$$
Using formulas
$$2\cdot \sin\alpha \cdot \sin\beta = \cos(\alpha-\beta) - \cos(\alpha+\beta),$$
$$2\cdot \sin\alpha \cdot \cos\beta = \sin(\alpha-\beta) + \sin(\alpha+\beta),$$
$$\sin(180^\circ-\varphi)  = \sin(\varphi), \quad \cos(180^\circ-\varphi)  = -\cos(\varphi),$$
we have
$$\sin 8^\circ \cdot (\cos 38^\circ - \cos 62^\circ) \mathop{=}\limits^{???} \sin 2^\circ \cdot (\cos 16^\circ + \cos 72^\circ),$$
$$-\sin 30^\circ + \sin 46^\circ  + \sin 54^\circ - \sin 70^\circ   \mathop{=}\limits^{???} -\sin 14^\circ + \sin 18^\circ - \sin 70^\circ + \sin 74^\circ,$$
$$-\sin 30^\circ + \sin 46^\circ  + \sin 54^\circ  + \sin 14^\circ - \sin 18^\circ - \sin 74^\circ \mathop{=}\limits^{???} 0,$$
$$(\sin 14^\circ  + \sin 46^\circ - \sin 74^\circ) + ( - \sin 18^\circ + \sin 54^\circ -\frac{1}{2} ) \mathop{=}\limits^{???} 0,$$
$$(\sin 14^\circ  + \sin 134^\circ + \sin 254^\circ) + \frac{1}{2}( - \sin 18^\circ + \sin 54^\circ +\sin 126^\circ + \sin 198^\circ + \sin270^\circ ) \mathop{=}\limits^{???} 0,$$
it is equality, because left side is equal to
$$\mathrm{Im} \left( p(\omega_3^0  + \omega_3^1 + \omega_3^2) + \frac{q}{2}( \omega_5^0 + \omega_5^1 +\omega_5^2 + \omega_5^3 + \omega_5^4) \right)=$$
$$\mathrm{Im} \left( p \cdot \frac{\omega_3^3-1}{\omega_3-1} + \frac{q}{2}\cdot \frac{\omega_5^5-1}{\omega_5-1} \right)=
\mathrm{Im} \left( p \cdot \frac{1-1}{\omega_3-1} + \frac{q}{2}\cdot \frac{1-1}{\omega_5-1} \right)=
\mathrm{Im} \left( p \cdot 0 + \frac{q}{2} \cdot 0 \right)= 0,$$
where
$p = \exp(i\pi 7/90)$, $q = \exp(-i\pi /10)$,
$w_{...}$ $-$ roots of unity:
$w_3 = \exp(i 2 \pi / 3) = \cos 2\pi/3 + i \sin 2\pi/3 = \cos 120^\circ + i \sin 120^\circ$, 
$w_5 = \exp(i 2 \pi / 5) = \cos 2\pi/5 + i \sin 2\pi/5 = \cos 72^\circ + i \sin 72^\circ$.
