13) In the adjacent figure, calculate $CD$ knowing that $OB=1$, $A \widehat{B} C=45^{\circ}, B \widehat{C} D=15^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_4c97eba1baa5e71bd15fg-2.jpg?height=351&width=328&top_left_y=414&top_left_x=1104)
13) The answer is (D). We notice that $D \widehat{A} B=D \widehat{C} B=15^{\circ}$, because both subtend the same arc. If we call $C^{\prime}$ the symmetric point of $C$ with respect to $O$ (which obviously lies on the circumference), we will have that $D \widehat{C^{\prime}} C=D \widehat{A C}=O \widehat{A} C-D \widehat{A} B=45^{\circ}-15^{\circ}=30^{\circ}$. Therefore, $C D C^{\prime}$ is a right triangle with a $30^{\circ}$ angle and we can conclude that the shorter leg $C D$ is half the hypotenuse $C C^{\prime}$. Since $C C^{\prime}=2 O B=2$, we have $C D=1$.

## Second Solution

Since $A \widehat{B} C=45^{\circ}$, we have $A \widehat{O} C=90^{\circ}$ and thus $B \widehat{O} C=90^{\circ}$ as well.

Since $B \widehat{C} D=15^{\circ}$, we have $D \widehat{O} B=30^{\circ}$ and thus $D \widehat{O} C=B \widehat{O} C-B \widehat{O} D=60^{\circ}$. $C D$ is therefore the side of the inscribed regular hexagon in the circle and its length is therefore equal to 1.