5. Given that in isosceles $\triangle A B C$ there is a point $K$, such that $A B=B C=C K, \angle K A C=30^{\circ}$. Try to find $\angle A K B$.

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The translation is provided as requested, maintaining the original format and line breaks.
5. $150^{\circ}$.

Solution 1 As shown in Figure 2, construct an equilateral $\triangle A C L$ with $A C$ as a side, where points $L$ and $B$ are on the same side of line $A C$.

In $\triangle A B C$, draw the altitude $B M$, which is easily known to be the perpendicular bisector of segment $A C$.
Since $\triangle A C L$ is an equilateral triangle, point $L$ is also on line $B M$.

Similarly, in $\triangle A C L$, draw the altitude $A N$, and we know that point $K$ is on line $A N$.

Notice that points $K$ and $A$ are on the same side of $B M$, and $C K = C B$.
Thus, point $K$ is inside $\triangle B M A$.
Therefore, point $K$ is on segment $A N$.
By $M C = \frac{A C}{2} = \frac{L C}{2} = N C$ and $B C = K C$, we know
Rt $\triangle B M C \cong$ Rt $\triangle K N C$.
Thus, $B M = K N$.
Also, $B M = K N = A B = B C$.
Therefore, in $\triangle A B C$,
$\angle A B C > 60^{\circ} > \angle B A C$.
Thus, points $P$ and $B$ are on opposite sides of line $A C$.
Since $\triangle A B C$ and $\triangle P B C$ are both isosceles triangles, we have
$\angle B C P = \frac{180^{\circ} - \angle P B C}{2}$,
$\angle B C A = \frac{180^{\circ} - \angle A B C}{2}$.
Also, since $\triangle A B P$ is an isosceles triangle, we get
$\angle P C A = \angle P C B - \angle A C B$
$= \frac{(180^{\circ} - \angle P B C) - (180^{\circ} - \angle A B C)}{2}$
$= \frac{\angle A B C - \angle P B C}{2} = \frac{\angle A B P}{2} = 30^{\circ}$
$\Rightarrow \angle K A C = 30^{\circ} = \angle A C P$
$\Rightarrow A K \parallel C P$.
Given $K C = A P$, quadrilateral $A K C P$ is either a parallelogram or an isosceles trapezoid.
Notice that the base angles of an isosceles triangle must be acute.
Thus, $\angle K A P < \angle B A P = 60^{\circ} < 90^{\circ}$,
$\angle K C P < \angle P C B < 90^{\circ}$.
And $\angle K A P + \angle K C P < 180^{\circ}$, which contradicts the fact that the opposite angles of an isosceles trapezoid sum to $180^{\circ}$.
Therefore, quadrilateral $A K C P$ must be a parallelogram.
Hence, $\angle K C A = \angle C A P$.
Now, let's calculate the angles.
Let $\angle B A K = \alpha$, then
$\angle C A P = \angle B A P - \angle B A K - \angle K A C$
$= 30^{\circ} - \alpha$
Notice that,
$\angle B C K = \angle B C A - \angle K C A$
$= \angle B A C - \angle K C A$
$= \alpha + 30^{\circ} - (30^{\circ} - \alpha) = 2 \alpha$
Thus, in $\triangle B K C$, we have
$\angle K B C = \frac{180^{\circ} - \angle B C K}{2}$
$= \frac{180^{\circ} - 2 \alpha}{2} = 90^{\circ} - \alpha$.
And in $\triangle A B C$, we have
$\angle A B C = 180^{\circ} - 2 \angle B A C$
$= 180^{\circ} - 2(30^{\circ} + \alpha) = 120^{\circ} - 2 \alpha$.
Consider $\triangle A B K$.
By $\angle A B K = 30^{\circ} - \alpha$ and $\angle B A K = \alpha$, we know
$\angle A K B = 150^{\circ}$.
Solution 3 As shown in Figure 4, in $\triangle A B C$, draw the altitude $B M$, and let it intersect $A K$ at point $O$.
Since $\angle A O M = 60^{\circ}$, and $\triangle A O C$ is symmetric about $B M$, we know $\angle C O M = 60^{\circ}$.
Thus, $\angle A O C = 120^{\circ}$, $\angle C O B = 120^{\circ}$.
This indicates that rays $O B$ and $O K$ are symmetric about line $C O$.
Since $B$ and $K$ are the intersections of these two rays with a circle centered at $C$ (which is also symmetric about $C O$), points $B$ and $K$ are also symmetric. In particular, $O B = O K$.
Thus, $\triangle B K O$ is an isosceles triangle with an internal angle of $120^{\circ}$ at vertex $O$.
Therefore, $\angle O K B = 30^{\circ}$, $\angle A K B = 150^{\circ}$.