12.181. 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 $\alpha$.
## Solution.

Let $A C=B C=1, D C=x$ (Fig. 12.50).

Then $C E=3 x$.

If $L$ is the midpoint of $D E$, then $O C \perp A B, O L \perp D E, \angle C O L=\angle A C D$ as acute angles with mutually perpendicular sides, $D L=E L=2 x$, $C L=x$. Since $D C \cdot C E=A C \cdot B C$, then $x \cdot 3 x=1 \cdot 1 ; x=\frac{1}{\sqrt{3}}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0933.jpg?height=422&width=458&top_left_y=96&top_left_x=106)

Fig. 12.50

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0933.jpg?height=438&width=656&top_left_y=78&top_left_x=607)

Fig. 12.51

$\angle A O C=\frac{1}{2} \angle A O B=\frac{\alpha}{2}$ and from $\triangle A C O\left(\angle A C O=90^{\circ}\right):$

$O C=A C \operatorname{ctg} \angle A O C=\operatorname{ctg} \frac{\alpha}{2}$.

From $\triangle O L C\left(\angle O L C=90^{\circ}\right): \sin \angle C O L=\frac{C L}{O C}=\frac{1}{\sqrt{3} \operatorname{ctg} \frac{\alpha}{2}}=\frac{\operatorname{tg} \frac{\alpha}{2}}{\sqrt{3}}$.

Thus, $\angle A C D=\angle C O L=\arcsin \frac{\operatorname{tg} \frac{\alpha}{2}}{\sqrt{3}}$.

Since $\sin \angle C O L \leq 1$, then $\frac{\operatorname{tg} \frac{\alpha}{2}}{\sqrt{3}} \leq 1 ; \operatorname{tg} \frac{\alpha}{2} \leq \sqrt{3} ; \alpha \leq 120^{\circ}$.

Answer: $\arcsin \frac{\operatorname{tg} \frac{\alpha}{2}}{\sqrt{3}}, \alpha \leq 120^{\circ}$.