12.190. A line is drawn through the vertex of an equilateral triangle, dividing the base in the ratio $2: 1$. At what angles is it inclined to the lateral sides of the triangle?
Solution.

The line $B D$ divides the side $A C$ of the equilateral triangle $A B C$ in the ratio $2: 1, C D: A D=2: 1$.

Let $A D=x, \angle A B D=\alpha$, then $C D=2 x, A B=3 x, \angle A D B=120^{\circ}-\alpha$. From $\triangle A D B$ :

$$
\begin{aligned}
& \frac{\sin \angle A D B}{\sin \angle A B D}=\frac{A B}{A D} \Rightarrow \frac{\sin \left(120^{\circ}-\alpha\right)}{\sin \alpha}=3 \Leftrightarrow \frac{\frac{\sqrt{3}}{2} \cos \alpha+\frac{1}{2} \sin \alpha}{\sin \alpha}=3 \Leftrightarrow \\
& \Leftrightarrow \frac{\sqrt{3}}{2} \operatorname{ctg} \alpha+\frac{1}{2}=3 \Leftrightarrow \operatorname{ctg} \alpha=\frac{5}{\sqrt{3}} \Leftrightarrow \operatorname{tg} \alpha=\frac{\sqrt{3}}{5} .
\end{aligned}
$$

Thus, $\angle A B D=\operatorname{arctg} \frac{\sqrt{3}}{5}, \angle C B D=60^{\circ}-\operatorname{arctg} \frac{\sqrt{3}}{5}$.

Answer: $\operatorname{arctg} \frac{\sqrt{3}}{5}, 60^{\circ}-\operatorname{arctg} \frac{\sqrt{3}}{5}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0940.jpg?height=484&width=732&top_left_y=75&top_left_x=288)

Fig. 12.58