Problem 4. Three sides of a quadrilateral are equal, and the angles of the quadrilateral formed by these sides are $90^{\circ}$ and $150^{\circ}$. Find the other two angles of this quadrilateral.
Answer: $45^{\circ}$ and $75^{\circ}$.

Solution. Let's denote the vertices of the quadrilateral as shown in the diagram.

Extend $A B C$ to form a square $A B C X$. In triangle $X C D$, the angle $\angle X C D$ is equal to $\angle B C D - \angle B C X = 150^{\circ} - 90^{\circ} = 60^{\circ}$, and the sides $C X$ and $C D$ are equal. Therefore, triangle $X C D$ is isosceles with an angle of $60^{\circ}$, meaning it is equilateral (in particular, segment $X D$ is also equal to the side of the square).

Now that we understand that

![](https://cdn.mathpix.com/cropped/2024_05_06_4fbead29144343cd679ag-2.jpg?height=250&width=372&top_left_y=1274&top_left_x=904)
our quadrilateral is formed by

![](https://cdn.mathpix.com/cropped/2024_05_06_4fbead29144343cd679ag-2.jpg?height=263&width=374&top_left_y=1548&top_left_x=903)
a square and an equilateral triangle, we can calculate its angles. Triangle $A X D$ is isosceles with an angle of $90^{\circ} + 60^{\circ} = 150^{\circ}$ at the vertex.

Therefore,

$$
\angle X A D = \angle X D A = \frac{180^{\circ} - 150^{\circ}}{2} = 15^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_4fbead29144343cd679ag-3.jpg?height=257&width=364&top_left_y=128&top_left_x=914)

Thus,

$$
\begin{aligned}
& \angle B A D = \angle B A X - \angle X A D = 90^{\circ} - 15^{\circ} = 75^{\circ} \\
& \angle A D C = \angle X D C - \angle X D A = 60^{\circ} - 15^{\circ} = 45^{\circ}
\end{aligned}
$$