12.189. A ray drawn from the vertex of an equilateral triangle divides its base in the ratio $m: n$. Find the obtuse angle between the ray and the base.
Solution.

Ray $B D$ divides side $A C$ of an equilateral triangle $A B C$ in the ratio $m: n$ (Fig. 12.57).

Let $\angle B D C=\alpha$ be the acute angle between the ray and the base, and assume that $n>m$. Then $\frac{A D}{D C}=\frac{m}{n}$.

If $A D=m x$, then $D C=n x, A B=$ $=A C=(m+n) x$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0938.jpg?height=511&width=521&top_left_y=1295&top_left_x=737)

Fig. 12.57
$\angle A=\frac{\pi}{3}, \angle A B D=\alpha-\frac{\pi}{3}$. By the Law of Sines, from $\triangle A D B$ :

$$
\begin{aligned}
& \frac{A B}{\sin \angle A D B}=\frac{A D}{\sin \angle A B D} \Rightarrow \frac{(m+n) x}{\sin (\pi-\alpha)}=\frac{m x}{\sin \left(\alpha-\frac{\pi}{3}\right)} \Leftrightarrow \frac{\sin \left(\alpha-\frac{\pi}{3}\right)}{\sin \alpha}= \\
& =\frac{m}{m+n} \Leftrightarrow \frac{\frac{1}{2} \sin \alpha-\frac{\sqrt{3}}{2} \cos \alpha}{\sin \alpha}=\frac{m}{m+n} \Leftrightarrow \frac{1}{2}-\frac{\sqrt{3}}{2} \operatorname{ctg} \alpha=\frac{m}{m+n} \Leftrightarrow \\
& \Leftrightarrow \sqrt{3} \operatorname{ctg} \alpha=1-\frac{2 m}{m+n}=\frac{n-m}{n+m} \Leftrightarrow \operatorname{tg} \alpha=\sqrt{3} \cdot \frac{n+m}{n-m}
\end{aligned}
$$

From this, $\alpha=\operatorname{arctg}\left(\sqrt{3} \cdot \frac{n+m}{n-m}\right)$ and the required obtuse angle is $\pi-\operatorname{arctg}\left(\sqrt{3} \cdot \frac{n+m}{n-m}\right)$.

Answer: $\pi-\operatorname{arctg}\left(\sqrt{3} \cdot \frac{n+m}{n-m}\right)$.