4. The angle at which the bisectors of two interior angles of a certain triangle intersect is $111^{\circ}$. One of them intersects with the third angle bisector at an angle of $122^{\circ}$. Determine the sizes of the interior angles of the triangle.
4. 

![](https://cdn.mathpix.com/cropped/2024_05_30_cbc9d69bfef3283ff9a2g-2.jpg?height=808&width=862&top_left_y=1618&top_left_x=583)

Sketch:

$1 \text{BOD}$

Let's assume, for example, that the angle bisectors drawn from vertices $A$ and $B$ intersect at an angle of $111^{\circ}$, and the angle bisectors drawn from vertices $A$ and $C$ intersect at an angle of $122^{\circ}$.

In triangle $ABD$, we have $\frac{\alpha}{2} + \frac{\beta}{2} + 111^{\circ} = 180^{\circ}$, which simplifies to $\frac{\alpha}{2} + \frac{\beta}{2} = 69^{\circ}$. Therefore, $\alpha + \beta = 138^{\circ}$. \quad $1 \text{BOD}$

| From triangle $ABC$, it follows that $\gamma = 180^{\circ} - 138^{\circ} = 42^{\circ}$. | $1 \text{BOD}$ |
| :---: | :---: |
| In triangle $ACD$, we have $\frac{\alpha}{2} + \frac{\gamma}{2} + 122^{\circ} = 180^{\circ}$, which simplifies to $\frac{\alpha}{2} + \frac{\gamma}{2} = 58^{\circ}$. Therefore, $\alpha + \gamma = 116^{\circ}$. | $1 \text{BOD}$ |
| From triangle $ABC$, it follows that $\beta = 180^{\circ} - 116^{\circ} = 64^{\circ}$ | $1 \text{BOD}$ |
| Then, $\alpha = 180^{\circ} - (42^{\circ} + 64^{\circ}) = 74^{\circ}$. The angles are $42^{\circ}$, $64^{\circ}$, and $74^{\circ}$. | $1 \text{BOD}$ |