1. **State the problem:** Find the smallest positive integer $x$ (in degrees) such that $$\tan(x - 160^\circ) = \frac{\cos 50^\circ}{1 - \sin 50^\circ}.$$\ 2. **Simplify the right side:** Recall the tangent half-angle identity: $$\tan \left( \frac{\theta}{2} \right) = \frac{\sin \theta}{1 + \cos \theta} = \frac{1 - \cos \theta}{\sin \theta}.$$ However, another useful form is: $$\frac{\cos \alpha}{1 - \sin \alpha} = \tan \left(45^\circ + \frac{\alpha}{2} \right)$$ because $$\tan \left(45^\circ + \frac{\alpha}{2} \right) = \frac{1 + \tan \frac{\alpha}{2}}{1 - \tan \frac{\alpha}{2}}$$ and using angle sum formulas leads to the given fraction. 3. **Apply the identity for $\alpha = 50^\circ$: ** $$\frac{\cos 50^\circ}{1 - \sin 50^\circ} = \tan \left(45^\circ + 25^\circ \right) = \tan 70^\circ.$$\ 4. **Rewrite the original equation:** $$\tan(x - 160^\circ) = \tan 70^\circ.$$\ 5. **Solve for $x$: ** Since $\tan \theta$ is periodic with period $180^\circ$, solutions satisfy $$x - 160^\circ = 70^\circ + k \times 180^\circ, \quad k \in \mathbb{Z}.$$\ 6. **Find the smallest positive solution:** $$x = 160^\circ + 70^\circ + k \times 180^\circ = 230^\circ + 180^\circ k.$$\ For $k = 0$, $x = 230^\circ$ (which is positive). For $k = -1$, $x = 50^\circ$ (also positive and smaller than 230°). Check $x = 50^\circ$: $$\tan(50^\circ - 160^\circ) = \tan(-110^\circ) = \tan(70^\circ)$$ (since $\tan(\theta + 180^\circ) = \tan \theta$). Indeed, $\tan(-110^\circ) = \tan(70^\circ)$ due to tangent periodicity. Thus, the smallest positive integer solution is: $$\boxed{50^\circ}.$$