Diameter $A B$ and chord $C D$ intersect at point $M, \angle C M B=73^{\circ}$, the angular magnitude of arc $B C$ is $110^{\circ}$. Find the magnitude of arc $B D$.
$C M B$ - the external angle of triangle $M D B$.

## Solution

Since $C M B$ is the external angle of triangle $M D B$, then

$$
\angle A B D=\angle C M B-\angle C D B=73^{\circ}-55^{\circ}=18^{\circ} .
$$

Therefore,

$$
\cup A D=36^{\circ}, \cup B D=180^{\circ}-36^{\circ}=144^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_1b17058527caf5b5d058g-04.jpg?height=472&width=548&top_left_y=667&top_left_x=755)

Answer

$144^{\circ}$.