6. An exterior angle and its adjacent interior angle of a triangle differ by $25^{\circ}$. The remaining two interior angles of the triangle also differ by $25^{\circ}$. Determine the measures of all the interior angles of the triangle. Examine all possibilities.
## 6. 1. case (5 POINTS):

Sketch:

![](https://cdn.mathpix.com/cropped/2024_05_30_59f5bb0a569372aa766ag-4.jpg?height=546&width=870&top_left_y=1600&top_left_x=593)

Let the external angle of the triangle, for example, $\beta_{1}$, be greater, so it holds that $\beta_{1}=\beta+25^{\circ}$.

1 POINT

Furthermore, it holds that

$\beta+\beta_{1}=180^{\circ}$

$\beta+\beta+25^{\circ}=180^{\circ}$

$2 \cdot \beta=180^{\circ}-25^{\circ}$

$\beta=155^{\circ}: 2$

$\beta=77^{\circ} 30'$

For the remaining two internal angles, the following equality holds:

$\alpha+\beta+\gamma=180^{\circ}$

$\alpha+77^{\circ} 30'+\alpha+25^{\circ}=180^{\circ}$

$2 \cdot \alpha=180^{\circ}-\left(77^{\circ} 30'+25^{\circ}\right)$

$2 \cdot \alpha=179^{\circ} 60'-102^{\circ} 30'$

$2 \cdot \alpha=77^{\circ} 30'$

$\alpha=76^{\circ} 90': 2$

$\alpha=38^{\circ} 45'$

2 POINTS

$\gamma=\alpha+25^{\circ}=38^{\circ} 45'+25^{\circ}=63^{\circ} 45'$