Consider the following figure with $O$ being the center of the circumcircle of $ABC$ (the circle passing through the three points $A, B$, and $C$). Additionally, we have $\widehat{A B C}=45^{\circ}$ and $\widehat{A C B}=62^{\circ}$. Find the measure of the angle $\widehat{B A O}$.

![](https://cdn.mathpix.com/cropped/2024_05_10_bca2b404556ff27bef16g-42.jpg?height=451&width=525&top_left_y=1159&top_left_x=731)
As the inscribed angle $\widehat{A C B}$ and the central angle $\widehat{A O B}$ intercept the same arc $A B$, we have $\widehat{A O B}=2 \widehat{A C B}=124^{\circ}$. Furthermore, the triangle $A O B$ is isosceles (since $O$ is the center of the circumscribed circle) so $\widehat{B A O}=\widehat{A B O}=\frac{180^{\circ}-124^{\circ}}{2}=28^{\circ}$.