In an isosceles trapezoid $A B C D(A D \| B C)$, the distance from vertex $A$ to the line $C D$ is equal to the lateral side. Find the angles of the trapezoid if $A D: B C=5$.
Let $M$ be the projection of vertex $A$ onto $CD$, and $K$ be the projection of $C$ onto $AD$. Denote $BC = x$, $AD = 5x$, $AM = CD = y$. Then $DK = \frac{1}{2}(AD - BC) = 2x$.

From the similarity of triangles $CKD$ and $AMD$, it follows that $CK : CD = AM : AD$, that is, $\frac{\sqrt{y^2 - 4x^2}}{y} = y / 5x$, or $\sqrt{1 - 4\left(\frac{x}{y}\right)^2} = y / 5x$.

Solving this equation for $(x / y)^2$, we get $(x / y)^2 = 1 / 5$ or $(x / y)^2 = 1 / 20$.

In the first case, $y = x \sqrt{5}$, $\cos \angle CDA = KD / CD = \frac{2}{\sqrt{5}}$.

In the second case, $-\cos \angle CDA = KD / CD = \frac{1}{\sqrt{5}}$.

## Answer

$\arccos \frac{1}{\sqrt{5}}, \pi - \arccos \frac{1}{\sqrt{5}}$ or $\arccos \frac{2}{\sqrt{5}}, \pi - \arccos \frac{2}{\sqrt{5}}$.

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