[ Application of trigonometric formulas (geometry).]

On the extension of side $A D$ of rectangle $A B C D$ beyond point $D$, a point $E$ is taken such that $D E=0.5 A D, \angle B E C=$ $30^{\circ}$.

Find the ratio of the sides of rectangle $A B C D$.
Apply the formula for the tangent of the difference.

## Solution

Let $\angle B E A=\alpha, \angle C E A=\beta, AB / BC=t$. Then $\operatorname{tg} \alpha=2 t / 3, \operatorname{tg} \beta=2 t, \operatorname{tg} 30^{\circ}=\operatorname{tg}(\beta-\alpha)=\frac{\operatorname{tg} \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} \alpha \operatorname{tg} \beta}$.

Substituting the found expressions for $\operatorname{tg} \alpha$ and $\operatorname{tg} \beta$, after simplification we get $4 t^{2}-4 \sqrt{3} t+3=0$, which is $(2 t-\sqrt{3})^{2}$, hence $t=\frac{\sqrt{3}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_89dcdfe8766a8b2a52b5g-03.jpg?height=518&width=723&top_left_y=404&top_left_x=679)

## Answer

$\sqrt{3}: 2$.

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