12.148. The ratio of the area of rectangle $ABCD (BC \| AD)$ to the square of its diagonal is $k$. Find $\angle EAF$, where $E$ and $\boldsymbol{F}$ are the midpoints of sides $BC$ and $CD$, respectively.
Solution.

Let $AB = y, AD = x, \angle BAE = \alpha, \angle DAF = \beta, \angle EAF = \gamma$ (Fig. 12.21). The diagonal of the rectangle is $\sqrt{x^2 + y^2}$, the area is $xy$, and $\frac{xy}{x^2 + y^2} = k \Rightarrow$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0907.jpg?height=363&width=567&top_left_y=172&top_left_x=112)

Fig. 12.21

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0907.jpg?height=456&width=454&top_left_y=77&top_left_x=774)

Fig. 12.22

\[
\begin{aligned}
& \Rightarrow \tan \gamma = \tan \left(90^\circ - (\alpha + \beta)\right) = \frac{1}{\tan (\alpha + \beta)} = \frac{1 - \tan \alpha \tan \beta}{\tan \alpha + \tan \beta} = \frac{1 - \frac{x}{2y} \cdot \frac{y}{2x}}{\frac{x}{2y} + \frac{y}{2x}} = \\
& = \left(1 - \frac{1}{4}\right) : \frac{x^2 + y^2}{2xy} = \frac{3xy}{2(x^2 + y^2)} = \frac{3k}{2} ; \gamma = \arctan \frac{3k}{2}
\end{aligned}
\]

Answer: $\arctan \frac{3k}{2}$.