# 3. Let the intersection point of segments

SK and AL be O. Note that CO is the median drawn to the hypotenuse of the right triangle ACL. Therefore, $\mathrm{AO}=\mathrm{OC}=\mathrm{OL}$ and

$\angle \mathrm{OCA}=\angle \mathrm{OAC}=\angle \mathrm{OAK}$

![](https://cdn.mathpix.com/cropped/2024_05_06_ed49bbf28b912afcdae4g-1.jpg?height=331&width=532&top_left_y=1576&top_left_x=1453)

(the last equality holds because AO is the angle bisector). Let this angle be $\alpha$.

We will find angle B. Since triangle CBK is isosceles, this angle is equal to the exterior angle of triangle CKA at B, that is, $\angle \mathrm{B}=\angle \mathrm{ACK}+\angle \mathrm{KAC}=3 \alpha$. From the equality $\angle \mathrm{B}+\angle \mathrm{A}=90^{\circ}$, we get that $2 \alpha+3 \alpha=90^{\circ}$. Therefore, $\alpha=18^{\circ}$. Accordingly, $\angle \mathrm{B}=54^{\circ}$, and $\angle \mathrm{A}=36^{\circ}$.
Answer: $36^{\circ}, 54^{\circ}, 90^{\circ}$