12.156. The angle at vertex $A$ of trapezoid $A B C D$ is $\alpha$. The lateral side $A B$ is twice the length of the shorter base $B C$. Find the angle $B A C$.
Solution. Let $\angle B A C=\beta$ (Fig. 12.29).

Then $\angle B C A=\angle C A D=\alpha-\beta$. From $\triangle A B C: \frac{\sin \beta}{\sin (\alpha-\beta)}=\frac{B C}{A B}=\frac{1}{2} \Leftrightarrow$

$\Leftrightarrow 2 \sin \beta=\sin \alpha \cos \beta-\sin \beta \cos \alpha \Leftrightarrow 2=\sin \alpha \operatorname{ctg} \beta-\cos \alpha \Leftrightarrow$

$\Leftrightarrow \frac{\sin \alpha}{\operatorname{tg} \beta}=2+\cos \alpha ; \beta=\operatorname{arctg} \frac{\sin \alpha}{2+\cos \alpha}$.

Answer: $\operatorname{arctg} \frac{\sin \alpha}{2+\cos \alpha}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0914.jpg?height=462&width=652&top_left_y=78&top_left_x=334)

Fig. 12.30