8,9 |

Points $A$ and $B$ are connected by two arcs of circles, convex in opposite directions: $\cup A C B=$ $117^{\circ} 23^{\prime}$ and $\cup A D B=42^{\circ} 37^{\prime}$. The midpoints $C$ and $D$ of these arcs are connected to point $A$. Find the angle $C A D$.
Ray $A B$ passes between the sides of angle $C A D$. Therefore, $\angle C A D=\angle C A B+\angle B A D$.

## Solution

Since ray $A B$ passes between the sides of angle $C A D$, then

$$
\begin{gathered}
\angle C A D=\angle C A B+\angle B A D=\frac{1}{2} \cup B C+\frac{1}{2} \cup B D= \\
=\frac{\frac{1}{2} \cup A C B+\frac{1}{2} \cup A D B}{2}=\frac{\cup A C B+\cup A D B}{4}=40^{\circ}
\end{gathered}
$$

## Answer

$40^{\circ}$.