[ Law of Sines [ Inscribed angle is half of the central angle ]

## Diameter $A B$ of the circle is extended beyond point $B$ and a point $C$ is marked on the extension. From point $C$,

 a secant is drawn at an angle of $7^{\circ}$ to $A C$, intersecting the circle at points $D$ and $E$, counting from point $C$. It is known that $D C = 3$, and the angle $D A C$ is $30^{\circ}$. Find the diameter of the circle.
#
Apply the Law of Sines to triangle DOC ($O$ - the center of the circle).

## Solution

Let $O$ be the center of the circle. Then $\angle D O B=60^{\circ}$. By the Law of Sines in triangle $D O C$, we find that

$$
\frac{D C}{\sin \angle D O C}=\frac{O D}{\sin \angle D O C}, \text{ or } \frac{3}{\sin 60^{\circ}}=\frac{O D}{\sin 7^{\circ}}.
$$

Therefore,

$$
O D=2 \sqrt{3} \sin 7^{\circ}, A B=2 O D=4 \sqrt{3} \sin 7^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-38.jpg?height=386&width=986&top_left_y=333&top_left_x=542)

# Answer 

$4 \sqrt{3} \sin 7^{\circ}$.