## Task 3 - 170723

In an isosceles triangle, it is only known that the sum of the sizes of two interior angles and one exterior angle is exactly $300^{\circ}$. However, it is not specified which of the mentioned interior angles are base angles and whether the mentioned exterior angle belongs to one of these interior angles or not.

Determine all possibilities for the sizes of the three interior angles of this triangle!
When an isosceles triangle has the required property and \( x \) is the size of a base angle, \( y \) is the size of the angle at the vertex, and \( x' \) and \( y' \) are the sizes of the exterior angles (adjacent angles) corresponding to \( x \) and \( y \), respectively, then one of the equations

\[
\begin{array}{lll}
x + x + x' = 300^\circ & (1) & x + x + y' = 300^\circ \\
x + y + x' = 300^\circ & (3) & x + y + y' = 300^\circ
\end{array}
\]

holds. Since \( x' + x = 180^\circ \) and \( y' + y = 180^\circ \), it follows from both (1) and (4) that \( x = 120^\circ > 90^\circ \), which contradicts the fact that \( x \) is the size of a base angle.

From (2), since the exterior angle theorem gives \( y' = 2x \), we get \( 4x = 300^\circ \) and thus \( x = 75^\circ \) and \( y = 30^\circ \). Since \( x + x' = 180^\circ \), it follows from (3) that \( y = 120^\circ \) and \( x = 30^\circ \).

The only remaining possibilities for the sizes of the interior angles of this triangle are \( 75^\circ; 75^\circ; 30^\circ \) or \( 30^\circ; 30^\circ; 120^\circ \).

These sizes satisfy the requirements of the problem; because \( 75^\circ + 75^\circ + 30^\circ = 180^\circ \) and \( 30^\circ + 30^\circ + 120^\circ = 180^\circ \), they are the sizes of the interior angles of triangles, and in the first case, the sum of the sizes of the two interior angles and the exterior angle is \( 75^\circ + 75^\circ + (180^\circ - 30^\circ) = 75^\circ + 75^\circ + 150^\circ = 300^\circ \), and in the second case, \( 30^\circ + 120^\circ + (180^\circ - 30^\circ) = 30^\circ + 120^\circ + 150^\circ = 300^\circ \).