5. Let's introduce the notation: $\measuredangle A B C=2 \theta, \quad \measuredangle B C A=2 \beta$, $\measuredangle C A B=2 \varphi$. By the theorem of the sum of angles in triangle $B O A$ we get

![](https://cdn.mathpix.com/cropped/2024_05_06_45e1176a268d466d045fg-17.jpg?height=371&width=474&top_left_y=390&top_left_x=498)

$$
\measuredangle B O A=180^{\circ}-\theta-\varphi=180^{\circ}-\frac{180^{\circ}-2 \beta}{2}=90^{\circ}+\beta .
$$

By the condition, quadrilateral $E O D C$ is inscribed, therefore,

$$
\angle D O E=180^{\circ}-2 \beta .
$$

From the condition $\measuredangle B O A=\measuredangle D O E$ (vertical angles) we get the equation

$$
90^{\circ}+\beta=180^{\circ}-2 \beta
$$

From this, we find $\beta=30^{\circ}$. Therefore, $\measuredangle B C A=2 \beta=60^{\circ}$. In triangle $O D C \quad \angle D O C=58^{\circ}$ by the condition, $\measuredangle D C O=\beta=30^{\circ}$, since $O C$ is the bisector of angle $\measuredangle B C A$. Then

$$
\measuredangle O D C=180^{\circ}-\angle D O C-\angle D C O=180^{\circ}-58^{\circ}-30^{\circ}=92^{\circ} .
$$

Let's find $\angle D A C$ of triangle $D A C$

$$
\angle D A C=\varphi=180^{\circ}-\angle O D C-\angle D C A=180^{\circ}-92^{\circ}-60^{\circ}=28^{\circ} .
$$

Therefore, $\measuredangle C A B=2 \varphi=56^{\circ}$. It remains to find $\measuredangle A B C$ :

$$
\measuredangle A B C=180^{\circ}-\measuredangle C A B-\measuredangle B C A=180^{\circ}-56^{\circ}-60^{\circ}=64^{\circ} .
$$
Answer: $56^{\circ}, 64^{\circ}, 60^{\circ}$.