(F.Nilov) Given right triangle $ ABC$ with hypothenuse $ AC$ and $ \angle A \equal{} 50^{\circ}$. Points $ K$ and $ L$ on the cathetus $ BC$ are such that $ \angle KAC \equal{} \angle LAB \equal{} 10^{\circ}$. Determine the ratio $ CK/LB$.
1. **Given Information and Setup:**
   - We have a right triangle \( \triangle ABC \) with \( \angle A = 50^\circ \) and hypotenuse \( AC \).
   - Points \( K \) and \( L \) are on the cathetus \( BC \) such that \( \angle KAC = \angle LAB = 10^\circ \).
   - We need to determine the ratio \( \frac{CK}{LB} \).

2. **Using the Sine Rule in \( \triangle ALB \):**
   - In \( \triangle ALB \), we apply the sine rule:
     \[
     \frac{LB}{\sin(10^\circ)} = \frac{AB}{\sin(80^\circ)}
     \]
   - Since \( \sin(80^\circ) = \cos(10^\circ) \), we have:
     \[
     \frac{LB}{\sin(10^\circ)} = \frac{AB}{\cos(10^\circ)}
     \]
   - Therefore:
     \[
     LB = \frac{AB \cdot \sin(10^\circ)}{\cos(10^\circ)}
     \]

3. **Using the Sine Rule in \( \triangle KCA \):**
   - In \( \triangle KCA \), we apply the sine rule:
     \[
     \frac{KC}{\sin(10^\circ)} = \frac{AC}{\sin(50^\circ)}
     \]
   - We know that \( AC = \frac{AB}{\sin(40^\circ)} \) because \( \angle B = 40^\circ \) (since \( \angle A = 50^\circ \) and \( \angle C = 90^\circ \)).
   - Substituting \( AC \) in the sine rule:
     \[
     \frac{KC}{\sin(10^\circ)} = \frac{\frac{AB}{\sin(40^\circ)}}{\sin(50^\circ)}
     \]
   - Simplifying, we get:
     \[
     \frac{KC}{\sin(10^\circ)} = \frac{AB}{\sin(40^\circ) \cdot \sin(50^\circ)}
     \]
   - Using the identity \( \sin(40^\circ) \cdot \sin(50^\circ) = \frac{1}{2} \left( \cos(10^\circ) - \cos(90^\circ) \right) = \frac{1}{2} \cos(10^\circ) \):
     \[
     \frac{KC}{\sin(10^\circ)} = \frac{AB}{\frac{1}{2} \cos(10^\circ)} = \frac{2AB}{\cos(10^\circ)}
     \]
   - Therefore:
     \[
     KC = \frac{2AB \cdot \sin(10^\circ)}{\cos(10^\circ)}
     \]

4. **Comparing \( KC \) and \( LB \):**
   - From the previous steps, we have:
     \[
     LB = \frac{AB \cdot \sin(10^\circ)}{\cos(10^\circ)}
     \]
     \[
     KC = \frac{2AB \cdot \sin(10^\circ)}{\cos(10^\circ)}
     \]
   - Therefore:
     \[
     KC = 2 \cdot LB
     \]
   - Hence, the ratio \( \frac{CK}{LB} \) is:
     \[
     \frac{CK}{LB} = 2
     \]

The final answer is \(\boxed{2}\)