Let $C$ and $D$ be points inside angle $\angle AOB$ such that $5\angle COD = 4\angle AOC$ and  $3\angle COD = 2\angle DOB$. If $\angle AOB = 105^{\circ}$, find $\angle COD$
1. Given the relationships:
   \[
   5\angle COD = 4\angle AOC \quad \text{and} \quad 3\angle COD = 2\angle DOB
   \]
   we can express \(\angle AOC\) and \(\angle DOB\) in terms of \(\angle COD\):
   \[
   \angle AOC = \frac{5}{4} \cdot \angle COD
   \]
   \[
   \angle DOB = \frac{3}{2} \cdot \angle COD
   \]

2. Since \(\angle AOB = 105^\circ\), we know that:
   \[
   \angle AOB = \angle AOC + \angle COD + \angle DOB
   \]

3. Substitute the expressions for \(\angle AOC\) and \(\angle DOB\) into the equation:
   \[
   105^\circ = \frac{5}{4} \cdot \angle COD + \angle COD + \frac{3}{2} \cdot \angle COD
   \]

4. Combine the terms on the right-hand side:
   \[
   105^\circ = \left( \frac{5}{4} + 1 + \frac{3}{2} \right) \cdot \angle COD
   \]

5. Convert all terms to a common denominator (4):
   \[
   \frac{5}{4} + 1 + \frac{3}{2} = \frac{5}{4} + \frac{4}{4} + \frac{6}{4} = \frac{15}{4}
   \]

6. Thus, the equation becomes:
   \[
   105^\circ = \frac{15}{4} \cdot \angle COD
   \]

7. Solve for \(\angle COD\):
   \[
   \angle COD = 105^\circ \cdot \frac{4}{15} = 28^\circ
   \]

The final answer is \(\boxed{28^\circ}\)