Exercise 11. Let $A B C$ be a triangle such that $\widehat{B A C}=48^{\circ}$ and $\widehat{C B A}=17^{\circ}$. The bisector of the angle $\widehat{B A C}$ intersects the segment $[B C]$ at point $D$. Determine the value in degrees of the angle $\widehat{C D A}$.
Solution to Exercise 11 (Draw a figure.) We have $\widehat{C D A}=180^{\circ}-\widehat{A D B}=\widehat{D B A}+$ $\widehat{B A D}=\widehat{C B A}+\frac{1}{2} \widehat{B A C}=17+24=41^{\circ}$