Given an isosceles trapezoid, a circle is inscribed in it and a circle is circumscribed around it.

The ratio of the height of the trapezoid to the radius of the circumscribed circle is $\sqrt{2 / 3}$. Find the angles of the trapezoid.
Express the diagonal of a trapezoid in two ways: through the radius of the circumscribed circle and the acute angle of the trapezoid; through the height and the acute angle of the trapezoid.

## Solution

Let $\alpha$ be the acute angle of trapezoid $ABCD$, $h$ be the height of the trapezoid, $R$ be the radius of the circumscribed circle, and $M$ be the projection of vertex $C$ of the smaller base $BC$ onto the larger base $AD$. Then
$AC = 2R \cdot \sin \alpha, AM = CD = CM / \sin \alpha$ and $AM^2 + MC^2 = AC^2$, or

$$
h^2 / \sin^2 \alpha + h^2 = 4R^2 \cdot \sin^2 \alpha
$$

Since $h / R = \sqrt{2 / 3}$, then

$$
2(1 / \sin^2 \alpha + 1) / 3 = 4 \cdot \sin^2 \alpha, \text{ or } 6 \cdot \sin^4 \alpha - \sin^2 \alpha - 1 = 0.
$$

From this, we find that $\sin^2 \alpha = 1 / 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-19.jpg?height=395&width=608&top_left_y=454&top_left_x=725)

Answer

$45^{\circ}, 135^{\circ}$.