This is a question which is related to the hardest easy questions. Note that the general solution belongs on math.se and is not solved in simple way. This question is a puzzle and you need to prove the answer.
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Although this puzzle is a math problem, it involves enough complex thought and nonconventional process in its solution to count as a puzzle, in my opinion.
In particular, it is a challenge to solve this problem without using trigonometry, as simple angle-chasing will get you nowhere. If you try and map out all possible linear equations of variables, you will find that $\angle AKL$ and $\angle BLK$ cannot be solved for. At first glance it appears that you will need to use scale measurements to solve this problem.
However, it is possible, after some thought. A solution that involves no trigonometry follows.
Rotate $\triangle ALK$ 30 degrees around point $K$, so that $A$ is mapped to $B$. Then $L$ is mapped to a point which we will call $M$, and a new triangle $\triangle BMK$ is created.
Note that since $LA = AB$, $MB = AB$ as well. Further note that since $\angle KAL = 15^\circ$, $\angle KBM = 15^\circ$ as well, so that $\angle ABM = \angle ABK - \angle KBM = 75^\circ - 15^\circ = 60^\circ$.
Since $AB$ and $BM$ are of equal length and $\angle ABM = 60^\circ$, $\triangle ABM$ must be an equilateral triangle, and $AM$ must be of the same length as $AB$ and $BM$.
Now, since $\triangle ABM$ is equilateral, it is also isosceles with $M$ as an apex. Since $\triangle ABK$ is also isosceles in this way, both $K$ and $M$ are perpendicular to $AB$ at its midpoint. So $KM$ is a perpendicular bisector of $AB$, and also an angle bisector of $\angle AKB$.
Therefore $\angle BKM = 15^\circ$, and we rotate the triangle back to find that $\angle AKL = 15^\circ$.