## Task B-3.3.

Points $A, B$ and $C$ lie on a circle with center at point $O$, as shown in the figure. Chord $\overline{A C}$ intersects radius $\overline{O B}$ at point $D$. Determine the measure of angle $\varangle B D A$ if $\varangle O A C=20^{\circ}, \varangle O B C=$ $43^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_9b9f9f32e16538559f3ag-15.jpg?height=340&width=328&top_left_y=1041&top_left_x=1298)
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_9b9f9f32e16538559f3ag-15.jpg?height=483&width=486&top_left_y=1549&top_left_x=705)

According to the theorem of the inscribed and central angles, with the labels as in the figure, $\delta=2 \varepsilon$.

Also,

From the sum of the angles in triangles $A O D$ and $B D C$, we get:

$$
\begin{gathered}
20^{\circ}+\delta+\theta=43^{\circ}+\frac{\delta}{2}+\theta \\
\frac{\delta}{2}=43^{\circ}-20^{\circ}=23^{\circ}
\end{gathered}
$$

$$
\delta=46^{\circ}
$$

Thus,

$$
\theta=180^{\circ}-\delta-20^{\circ}=180^{\circ}-46^{\circ}-20^{\circ}=114^{\circ}
$$

and

$$
\gamma=180^{\circ}-\theta=180^{\circ}-114^{\circ}=66^{\circ}
$$