Given is a isosceles triangle ABC so that AB=BC. Point K is in ABC, so that CK=AB=BC and <KAC=30°.Find <AKB=? 
1. **Identify the given information and draw the triangle:**
   - Triangle \( \triangle ABC \) is isosceles with \( AB = BC \).
   - Point \( K \) is inside \( \triangle ABC \) such that \( CK = AB = BC \).
   - \( \angle KAC = 30^\circ \).

2. **Determine the center \( O \) of the circumcircle of \( \triangle AKC \):**
   - Since \( \triangle AKC \) has \( CK = AB = BC \), \( \triangle AKC \) is isosceles with \( CK = AK \).
   - The center \( O \) of the circumcircle of \( \triangle AKC \) is equidistant from \( A \), \( K \), and \( C \).

3. **Analyze \( \triangle OKC \):**
   - Since \( O \) is the center of the circumcircle of \( \triangle AKC \), \( \triangle OKC \) is equilateral because \( OK = OC = CK \).

4. **Reflect \( B \) about \( AC \):**
   - Since \( O \) is the center of the circumcircle and \( \triangle OKC \) is equilateral, \( O \) is the reflection of \( B \) about \( AC \).

5. **Reflect \( K \) about \( AC \) to get point \( L \):**
   - Let \( L \) be the reflection of \( K \) about \( AC \).
   - Since \( \triangle AKL \) is equilateral, \( AK = KL = AL \).

6. **Analyze the relationship between \( B \) and \( L \):**
   - Since \( B \) and \( L \) are reflections of each other about \( AC \), \( BL = KO \).
   - Given \( KO = CK = AB \), we have \( BL = AB \).

7. **Determine \( \angle AKB \):**
   - Since \( \triangle AKL \) is equilateral, \( \angle AKL = 60^\circ \).
   - \( \angle AKB \) is the external angle to \( \triangle AKL \) at \( K \), which is \( 180^\circ - \angle AKL = 180^\circ - 60^\circ = 120^\circ \).

8. **Verify the final angle:**
   - Since \( \angle KAC = 30^\circ \) and \( \angle AKL = 60^\circ \), the total angle \( \angle AKB = 180^\circ - 30^\circ = 150^\circ \).

The final answer is \( \boxed{150^\circ} \).