3.65. The height of a triangle divides the angle of the triangle in the ratio $2: 1$, and the base - into segments, the ratio of which (larger to smaller) is $k$. Find the sine of the smaller angle at the base and the permissible values of $k$.
3.65. According to the condition, $B D \perp A C, D C: A D=k, D C>A D, \angle D B C > \angle D B A, \angle D B C: \angle D B A=2: 1$ (Fig. 3.67). Let $\angle D B A=\alpha$; then $\angle D B C=2 \alpha$ and $\angle C=90^{\circ}-2 \alpha2$. Therefore,

$$
\sin C=\cos 2 \alpha=\frac{1-\operatorname{tg}^{2} \alpha}{1+\operatorname{tg}^{2} \alpha}=\frac{1-\frac{k-2}{k}}{1+\frac{k-2}{k}}=\frac{1}{k-1}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-542.jpg?height=299&width=583&top_left_y=489&top_left_x=320)

Fig. 3.67