Notes
angle formed by two squares ii solution

Solution to the Angle Formed by Two Squares II Puzzle

Solution by Isosceles Triangles

As the squares are the same size, triangle $O B C$ is isosceles with $O B$ and $O C$ of the same length. Also triangle $O A C$ is isosceles, with $O A$ and $O C$ the same length. Let $a$ be angle $O \hat{C} A$ and let $b$ be angle $O \hat{C} B$. The requested angle is $b - a$.

As $O B C$ is isosceles, and the angles in a triangle add up to $180^\circ$, angle $B \hat{O} C$ is $180^\circ - 2 b$. Similarly, angle $A \hat{O} C$ is $180^\circ - 2 a$. Therefore, angle $A \hat{O} B$ is

It is also the corner of a square, so is $90^\circ$. Hence $b - a = 45^\circ$.

Solution by Angle at the Circumference is Half the Angle at the Centre

As the squares are the same size, the circle centred at $O$ that passes through $A$ also passes through $B$ and $C$. Then the requested angle is half of angle $O \hat{O} B$, which is $90^\circ$, since the angle at the circumference is half the angle at the centre.

Solution by Invariance Principle

By drawing the squares so that they are adjacent without overlapping, the requested angle can be seen to be $45^\circ$.