347. Point $A$ is located between two parallel lines at distances $a$ and $b$ from them. This point serves as the vertex of an angle equal to $\alpha$ for all possible triangles, the other two vertices of which lie on the given lines. Find the minimum value of the area of such triangles.
347. If one of the sides of the triangle passing through $A$ forms an angle $\varphi$ with a line perpendicular to the given parallel lines, then the other side forms an angle $180^{\circ}-\varphi-\alpha$; finding these sides, we get that the area of the triangle is $-\frac{a b \sin \alpha}{2 \cos \varphi \cos (\varphi+\alpha)}=-\frac{a b \sin \alpha}{\cos \alpha+\cos (\alpha+2 \varphi)}$. This expression is minimal if $\alpha+2 \varphi=180^{\circ}$. Answer: $S_{\min }=a b \operatorname{ctg} \frac{\alpha}{2}$.