## Task 2 - 180822

Determine the sizes of the interior angles of a triangle $ABC$, for which the following statement about its exterior angles is true:

One of the exterior angles at vertex $A$ is $16^{\circ}$ larger, one of the exterior angles at vertex $B$ is $49^{\circ}$ smaller than one of the exterior angles at $C$.
If the sizes of the interior and exterior angles of triangle $ABC$ at $A$ are denoted by $\alpha$ and $\alpha^{\prime}$, at $B$ by $\beta$ and $\beta^{\prime}$, and at $C$ by $\gamma$ and $\gamma^{\prime}$, then the relationships that are generally valid and assumed between them are described by

\[
\begin{aligned}
& \alpha^{\prime}=180^{\circ}-\alpha=\gamma^{\circ}+16^{\circ}=180^{\circ}-\gamma+16^{\circ} \\
& \beta^{\prime}=180^{\circ}-\beta=\gamma^{\circ}-49^{\circ}=180^{\circ}-\gamma-49^{\circ}
\end{aligned}
\]

from which it follows that $\alpha=\gamma-16^{\circ}$ and $\beta=\gamma+49^{\circ}$, and using the theorem of the sum of the interior angles in a triangle: $\alpha+\beta+\gamma=180^{\circ}=3 \gamma+33^{\circ}$.

From this, we obtain $\gamma=49^{\circ}$, $\alpha=49^{\circ}-16^{\circ}=33^{\circ}$, and $\beta=49^{\circ}+49^{\circ}=98^{\circ}$.

Indeed, a triangle $ABC$ exists because $49^{\circ}+33^{\circ}+98^{\circ}=180^{\circ}$, and it also has exterior angles of the following sizes:

at $C$ with $\gamma^{\prime}=180^{\circ}-49^{\circ}=131^{\circ}$,

at $A$ with $\alpha^{\prime}=180^{\circ}-33^{\circ}=147^{\circ}=131^{\circ}+16^{\circ}=\gamma^{\prime}+16^{\circ}$, and

at $B$ with $\beta^{\prime}=180^{\circ}-98^{\circ}=82^{\circ}=131^{\circ}-49^{\circ}=\gamma^{\prime}-49^{\circ}$.