In a triangle $ ABC,$ let $ D$ and $ E$ be the intersections of the bisectors of $ \angle ABC$ and $ \angle ACB$ with the sides $ AC,AB,$ respectively. Determine the angles $ \angle A,\angle B, \angle C$ if $ \angle BDE \equal{} 24 ^{\circ},$ $ \angle CED \equal{} 18 ^{\circ}.$
1. Let \( T \) be the intersection of \( BD \) and \( CE \). Since \( \angle BDE = 24^\circ \) and \( \angle CED = 18^\circ \), we can use the fact that the sum of the angles in a triangle is \( 180^\circ \) to find \( \angle BTC \).

2. Note that \( \angle BDT = \angle BDE = 24^\circ \) and \( \angle CDT = \angle CED = 18^\circ \). Therefore, \( \angle BTC = 180^\circ - (\angle BDT + \angle CDT) = 180^\circ - (24^\circ + 18^\circ) = 138^\circ \).

3. Since \( \angle TCB + \angle TBC = 42^\circ \), we have:
   \[
   \angle BAC = 180^\circ - \angle BTC = 180^\circ - 138^\circ = 42^\circ
   \]

4. Let \( M \) be the reflection of \( D \) across the line \( CE \) and let \( N \) be the reflection of \( E \) across \( BD \). Since \( BD \) and \( CE \) are bisectors of \( \angle C \) and \( \angle B \), it follows that the points \( M \) and \( N \) are located on the line \( BC \).

5. Let the line \( BD \) meet \( EM \) at \( R \). We need to find \( \angle ERB \). Since \( \angle ERB = \angle RED + \angle EDR \), and given that \( \angle RED = 60^\circ \) and \( \angle EDR = 60^\circ \), we have:
   \[
   \angle ERB = 60^\circ + 60^\circ = 120^\circ
   \]

6. Similarly, \( \angle BRN = 60^\circ \) and \( \angle MRN = 60^\circ \). Since \( \angle NDM = 24^\circ = \angle RDN \), we can conclude that \( N \) is the excenter of the triangle \( RDM \).

7. To find \( \angle DMC \), we use the fact that \( \angle DMC = \angle RMN \). Since \( \angle RMN = \frac{\angle DMC + \angle RMN}{2} = 54^\circ \), we have:
   \[
   \angle DMC = 54^\circ
   \]

8. Finally, we can determine the angles of the triangle \( ABC \):
   \[
   \angle A = 96^\circ, \quad \angle B = 12^\circ, \quad \angle C = 72^\circ
   \]

The final answer is \( \boxed{ \angle A = 96^\circ, \angle B = 12^\circ, \angle C = 72^\circ } \)