[ Inscribed angle, based on the diameter ] Sine Theorem

[ Sine Theorem $\quad$]

Diagonal $B D$ of quadrilateral $A B C D$ is the diameter of the circle circumscribed around this quadrilateral. Find diagonal $A C$, if $B D=2, A B=1, \angle A B D: \angle D B C=4: 3$.
Triangle $B A D$ is a right triangle.

## Solution

Let $\angle A B D=4 \boldsymbol{\alpha}$, then $\angle D B C=3 \alpha$. From the right triangle $A B D$ (angle $A$ is a right angle, as an inscribed angle subtending the diameter $B D$), we find that

$$
\cos \angle A B D=\frac{A B}{B D}=\frac{1}{2}
$$

Therefore,

$$
\angle A B D=4 \alpha=60^{\circ}, \alpha=15^{\circ} \text {. }
$$

Then

$$
\angle C B D=3 \alpha=45^{\circ}, \angle A B C=105^{\circ}
$$

Let $R$ be the radius of the circle. Since $2 R=B D=2$, then
$A C=2 R \sin \angle A B C=2 \sin 105^{\circ}=2 \sin \left(60^{\circ}+45^{\circ}\right)=$

$=2\left(\sin 60^{\circ} \cos 45^{\circ}+\cos 60^{\circ} \sin 45^{\circ}\right)=\frac{\sqrt{2}+\sqrt{6}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-42.jpg?height=475&width=531&top_left_y=274&top_left_x=772)

Answer

$\frac{\sqrt{2}+\sqrt{6}}{2}$.