1. **Problem Statement:** We analyze the first problem (top-left circle) with center $O$, diameter $AB$, and point $C$ on the circle. Given angle $\angle BAC = 47^\circ$, find angle $x = \angle CBA$. 2. **Key Properties:** - $AB$ is a diameter, so $\angle ACB = 90^\circ$ (angle in a semicircle). - Triangle $ABC$ is inscribed in the circle. - Sum of angles in triangle $ABC$ is $180^\circ$. 3. **Apply Triangle Angle Sum:** $$\angle BAC + \angle CBA + \angle ACB = 180^\circ$$ Substitute known values: $$47^\circ + x + 90^\circ = 180^\circ$$ 4. **Solve for $x$:** $$x = 180^\circ - 90^\circ - 47^\circ$$ $$x = 43^\circ$$ 5. **Summary:** - The angle at $B$ between $BC$ and $BA$ is $43^\circ$. Final answer: $\boxed{43^\circ}$