12.150. 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 $k$.
Solution.

Let $A D$ be the height of $\triangle A B C$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0908.jpg?height=416&width=568&top_left_y=70&top_left_x=692)

Fig. 12.23

(Fig. 12.23), $\angle C A D: \angle B A D=2: 1 \Rightarrow$ $\Rightarrow A C>A B, C D>B D, C D: B D=k$.

If $\angle B A D=\alpha$, then $\angle C A D=2 \alpha$. In $\triangle A D B\left(\angle A D B=90^{\circ}\right): B D=A D \operatorname{tg} \alpha$. From $\triangle A D C\left(\angle A D C=90^{\circ}\right): C D=A D \operatorname{tg} 2 \alpha$.

\[
\begin{aligned}
& \text { Then } k=\frac{C D}{B D}=\frac{A D \operatorname{tg} 2 \alpha}{A D \operatorname{tg} \alpha}=\frac{\operatorname{tg} 2 \alpha}{\operatorname{tg} \alpha}=\frac{2 \operatorname{tg} \alpha}{\operatorname{tg} \alpha\left(1-\operatorname{tg}^{2} \alpha\right)}=\frac{2}{1-\operatorname{tg}^{2} \alpha} \Rightarrow \\
& \Rightarrow \operatorname{tg}^{2} \alpha=\frac{k-2}{k} \Rightarrow k>2
\end{aligned}
\]

Further,

\[
\sin \angle C=\sin \left(90^{\circ}-2 \alpha\right)=\cos 2 \alpha=\frac{1-\operatorname{tg}^{2} \alpha}{1+\operatorname{tg}^{2} \alpha}=\frac{1-\frac{k-2}{k}}{1+\frac{k-2}{k}}=\frac{2}{2 k-2}=\frac{1}{k-1}
\]

Answer: $\frac{1}{k-1}, k>2$.