Volchkovich M.A.

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

![](https://cdn.mathpix.com/cropped/2024_05_06_8027e864227276d12944g-14.jpg?height=389&width=554&top_left_y=1695&top_left_x=752)

Extend \(ABC\) to form a square \(ABCX\). In triangle \(XCD\), the angle \(\angle XCD\) is \(\angle BCD - \angle BCX = 150^\circ - 90^\circ = 60^\circ\), and the sides \(CX\) and \(CD\) are equal. Therefore, triangle \(XCD\) is isosceles with an angle of \(60^\circ\), making it equilateral (in particular, segment \(XD\) is also equal to the side of the square).

![](https://cdn.mathpix.com/cropped/2024_05_06_8027e864227276d12944g-14.jpg?height=383&width=546&top_left_y=2264&top_left_x=756)

Now that we understand that our quadrilateral is formed from a square and an equilateral triangle, we can calculate its angles. Triangle \(AXD\) is isosceles with an angle of \(90^\circ + 60^\circ = 150^\circ\) at the vertex. Therefore, \(\angle XAD = \angle XDA = (180^\circ - 150^\circ) / 2 = 15^\circ\).

![](https://cdn.mathpix.com/cropped/2024_05_06_8027e864227276d12944g-15.jpg?height=383&width=560&top_left_y=0&top_left_x=752)

Thus,

\[
\angle BAD = \angle BAX - \angle XAD = 90^\circ - 15^\circ = 75^\circ;
\]

\[
\angle ADC = \angle XDC - \angle XDA = 60^\circ - 15^\circ = 45^\circ.
\]

## Answer

\(45^\circ\) and \(75^\circ\).