Problem 4. Two lines intersect at one point and form four angles (two acute and two obtuse). The sum of the opposite acute angles is equal to half of one of the opposite obtuse angles. Determine the sizes of the angles?
Solution. First method. Let the acute angles be equal to each other and let their size be $x$. Then, the obtuse angles have a size of $180^{\circ}-x$. From the condition, we have $2 x=\frac{1}{2}\left(180^{\circ}-x\right)$, which means $4 x=180^{\circ}-x$,
from which $5 x=180^{\circ}$, so $x=36^{\circ}$ is the size of the acute angles. (5) The obtuse angles have a size of $180^{\circ}-36^{\circ}=144^{\circ}$.

Second method. Let $\alpha$ be the acute angle and $\beta$ be the obtuse angle. From the condition of the problem, we have that the obtuse angle is twice the sum of the acute angles, i.e., $\beta=2 \cdot(\alpha+\alpha)=4 \alpha$ or the obtuse angle is four times larger than the acute angle. Therefore, the acute angle is $180^{\circ}: 5=36^{\circ}$, while the obtuse angle is $180^{\circ}-36^{\circ}=144^{\circ}$.