3.60. 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.
3.60. According to the condition, in an equilateral triangle $ABC$ we have $AL: LC = n: m$ (Fig. 3.62); it is required to find the angle $ALB$. Let $AC = b$, $AL = nx$, $LC = mx$; then $nx + mx = b$, from which $x = \frac{b}{m+n}$, $AL = \frac{nb}{m+n}$, $LC = \frac{mb}{m+n}$. Draw $BB_1 \perp AC$;

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-538.jpg?height=430&width=481&top_left_y=108&top_left_x=671)

Fig. 3.62

from this $\quad BB_1 = \frac{b \sqrt{3}}{2}$, $\quad B_1C = \frac{b}{2}$,

$LB_1 = \frac{mb}{m+n} - \frac{b}{2} = \frac{mb - nb}{2(m+n)} = \frac{(m-n)b}{2(m+n)}$. From $\triangle BB_1L$ we find

$$
\operatorname{tg} \angle BLB_1 = \frac{BB_1}{LB_1} = \frac{2b \sqrt{3}(m+n)}{2(m-n)b} = \frac{(m+n) \sqrt{3}}{m-n} \text{; }
$$

Thus, $\angle BLB_1 = \operatorname{arctg} \frac{(m+n) \sqrt{3}}{m-n}$, hence $\angle ALB = \pi - \operatorname{arctg} \frac{(m+n) \sqrt{3}}{m-n}$.