In a circle, it was possible to draw the radii $O A, O B$, and then the square $K L M N$, such that $K$ and $L$ bisect the drawn radii, and $M$ and $N$ are on the circumference of the circle. What is the measure of the angle $A O B$?
Let the measure of the angle sector $AOB$ containing points $M$ and $N$ be $2x$, the midpoints of sides $KL$ and $NM$ be $P$ and $Q$ respectively, and $OK=1$.
![](https://cdn.mathpix.com/cropped/2024_05_02_262b0b0ff329b9b86bfdg-1.jpg?height=382&width=1004&top_left_y=278&top_left_x=542)

Considering the direction of $OQ$ as positive, including the sign, we have $OP=\cos x$, $PQ=KN=KL=2\sin x$, and $MQ=\sin x$, since $OK=OL$ implies that $OP$ is the perpendicular bisector of segment $KL$ and the axis of symmetry of the figure, thus $\angle QOA = x$. Therefore, from the right triangle $OMQ$,

$$
(\cos x + 2 \sin x)^2 + \sin^2 x = 4
$$

It is easy to transition to functions of the unknown $2x$:

$$
\begin{gathered}
4 \sin x \cos x + 4 \sin^2 x = 2 \sin 2x + 2 - 2 \cos 2x = 3 \\
\sin 2x - \cos 2x = \frac{1}{2}
\end{gathered}
$$

Dividing the equation by $\sqrt{2}$ and recognizing the trigonometric functions of $45^\circ$ in the coefficients, based on the addition formula,

$$
\begin{gathered}
\sin \left(2x - 45^\circ\right) = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4} = 0.3536 \\
2x - 45^\circ = \left\{\begin{array}{c}
20^\circ 42' \\
159^\circ 18'
\end{array}\right.
\end{gathered}
$$

Thus, we obtained two values for the angle $AOB$: $65^\circ 42'$ and $204^\circ 18'$.

Remark. In solving (1), we followed the general solution of the equation

$$
a \sin x + b \cos x = c
$$

by introducing the auxiliary angle $\varphi$ such that

$$
\frac{a}{\sqrt{a^2 + b^2}} = \cos \varphi \quad \text{and} \quad \frac{b}{\sqrt{a^2 + b^2}} = \sin \varphi
$$

provided that $|c| \leq \sqrt{a^2 + b^2}$.