Notes
five equilateral triangles ii solution

Solution to the Five Equilateral Triangles II Puzzle

Solution by Rotation, Angles in Isosceles Triangles, Corresponding Angles, and Angles at a Point on a Straight Line

In the above diagram, triangle $A B F$ is another large equilateral triangle and $H I J$ is another small one.

Triangle $A G J$ is obtained from triangle $A B F$ by rotating $120^\circ$ about $A$. This also takes triangle $E D F$ to triangle $H I J$. Therefore, line segment $A I$ is the rotation by $120^\circ$ of line segment $A D$ about $A$. Hence angle $D \hat{A} I$ is $120^\circ$ and triangle $A D I$ is isosceles. This means that angle $I \hat{D} A$ is $30^\circ$ since angles in a triangle add up to $180^\circ$.

Finally, line segments $C J$ and $D I$ are parallel, so angles $J \hat{K} A$ and $I \hat{D} A$ are equal as they are corresponding angles. Then angle $A \hat{K} C = 180^\circ - 30^\circ = 150^\circ$ since angles at a point on a straight line add up to $180^\circ$.

Solution by Invariance Principle

The relative sizes of the equilateral triangles is not specified, and varying this leads to two configurations where the answer is more clear. In both, the lines have to be extended to ensure that the angle is well-defined.

In this configuration, all the equilateral triangles are the same size. The extra line bisects one of the equilateral triangles, meaning that the acute angle is half of $60^\circ$ and so the requested angle is $150^\circ$.

In this configuration, the smaller equilateral triangles have been shrunk down to a point. The requested angle is $150^\circ$ by a similar argument.