12.405 In an isosceles triangle $A B C$, an equilateral triangle $D E F$ is inscribed; point $D$ lies on side $B C$, point $E$ - on side $A C$, and point $F$ - on side $A B$. The side $A B$ is to the side $D F$ as 8:5. Find the sine of angle $D E C$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-290.jpg?height=295&width=391&top_left_y=494&top_left_x=90)

Let $\alpha=\angle D E C$. Then $\angle E D C=120^{\circ}-\alpha$, $\angle F D B=180^{\circ}-\left(60^{\circ}+\angle E D C\right)=\alpha \Rightarrow$ $\Rightarrow \triangle E D C=\triangle F B D, E C=B D$. By the Law of Sines for $\triangle E D C \frac{E C}{\sin \left(120^{\circ}-\alpha\right)}=\frac{D C}{\sin \alpha}=\frac{E D}{\sin 60^{\circ}} \Rightarrow$

$$
\Rightarrow B C=E C+D C=\frac{E D}{\sin 60^{\circ}}\left(\sin \left(120^{\circ}-\alpha\right)+\sin \alpha\right)=2 \cdot E D \cdot \cos \left(60^{\circ}-\alpha\right) \Rightarrow
$$

$$
\begin{aligned}
& \Rightarrow \cos \left(60^{\circ}-\alpha\right)=\frac{B C}{2 \cdot E D}=\frac{4}{5}, \sin \left(60^{\circ}-\alpha\right)= \pm \frac{3}{5} \Rightarrow \\
& \Rightarrow\left\{\begin{array}{l}
\cos \alpha+\sqrt{3} \cdot \sin \alpha=\frac{8}{5}, \\
\sqrt{3} \cdot \cos \alpha-\sin \alpha= \pm \frac{6}{5}
\end{array} \Rightarrow \sin \alpha=\frac{4 \sqrt{3} \pm 3}{10}\right.
\end{aligned}
$$

Answer: $\frac{4 \sqrt{3} \pm 3}{10}$.