3.72. The smaller arc of a circle subtended by the chord $A B$ contains $\alpha^{\circ}$. Through the midpoint $C$ of the chord $A B$, a chord $D E$ is drawn such that $D C: C E=1: 3$. Find the acute angle $A C D$ and the permissible values of $\alpha$.
3.72. By the condition, $AB$ is a chord, $\cup ANB=\alpha^{\circ}; AC=BC, DE$ is a chord, $DC:CE=1:3$ (Fig. 3.73). Let $CD=x$; then $EC=3x, ED=4x$. Draw $OF \perp ED$ and connect points $O$ and $C, O$ and $A$. Since $OC \perp AB$ (diameter $MN$ is perpendicular to $AB$), we have $\angle FOC=\angle ACD$ (acute angles with mutually perpendicular sides). Note that $FD=\frac{1}{2}ED=2x, FC=x$. Further, we have $BC \cdot AC=DC \cdot EC$, or $AC^2=x \cdot 3x$, i.e., $AC=x\sqrt{3}$. From $\triangle OCA$

we get $OC=AC \operatorname{ctg} \frac{\alpha}{2}=x\sqrt{3} \operatorname{ctg} \frac{\alpha}{2}$, and from $\triangle FOC$ we find

$$
\sin \angle FOC=\frac{FC}{OC}=\frac{x}{x\sqrt{3} \operatorname{ctg} \frac{\alpha}{2}}=\frac{\operatorname{tg} \frac{\alpha}{2}}{\sqrt{3}}
$$

Thus, $\angle ACD=\angle FOC=\arcsin \frac{\operatorname{tg} \frac{\alpha}{2}}{\sqrt{3}};$ since $\sin \angle ACD \leq 1$, then $\operatorname{tg} \frac{\alpha}{2} \leq \sqrt{3}$, i.e., $\alpha \leq 120^{\circ}$.