In the right-angled triangle $\mathrm{ABC}$, the angle at vertex $B$ is $30^{\circ}$. The center of the square constructed outward on the hypotenuse $\mathrm{ABC}$ is $D$. What is the measure of the angle $A D B$?
Let's denote the sought angle $A D B$ by $\delta$, then

$$
B A D \varangle=180^{\circ}-\left(75^{\circ}+\delta\right)=105^{\circ}-\delta
$$

We write the sine theorem in the triangle $A B D$:

$$
\begin{aligned}
\frac{\sin \delta}{\sin \left(105^{\circ}-\delta\right)} & =\frac{\sqrt{3}}{\sqrt{2}}, \quad \text { hence } \\
\sin \delta & =\frac{\sqrt{3}}{\sqrt{2}}\left(\sin 105^{\circ} \cos \delta-\cos 105^{\circ} \sin \delta\right)
\end{aligned}
$$

Using the angle sum formulas, we calculate the sine and cosine of $105^{\circ}$:

$$
\begin{aligned}
& \cos 105^{\circ}=\cos 60^{\circ} \cos 45^{\circ}-\sin 60^{\circ} \sin 45^{\circ}=\frac{1}{2} \frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2} \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}\left(\frac{1-\sqrt{3}}{2}\right) \\
& \sin 105^{\circ}=\sin 60^{\circ} \cos 45^{\circ}+\cos 60^{\circ} \sin 45^{\circ}=\frac{\sqrt{3}}{2} \frac{\sqrt{2}}{2}+\frac{1}{2} \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}\left(\frac{\sqrt{3}+1}{2}\right)
\end{aligned}
$$

Substitute the values obtained into equation (1):

$$
\begin{aligned}
& \sin \delta=\frac{\sqrt{3}}{\sqrt{2}}\left[\frac{\sqrt{2}}{2} \frac{\sqrt{3}+1}{2} \cos \delta-\frac{\sqrt{2}}{2} \frac{1-\sqrt{3}}{2} \sin \delta\right] \\
& \sin \delta=\frac{\sqrt{3}(\sqrt{3}+1)}{4} \cos \delta-\frac{\sqrt{3}(1-\sqrt{3})}{4} \sin \delta
\end{aligned}
$$

Divide by $\cos \delta \neq 0$:

$$
\operatorname{tg} \delta=\frac{3+\sqrt{3}}{4}-\frac{\sqrt{3}-3}{4} \operatorname{tg} \delta
$$

Thus,

$$
\begin{gathered}
\operatorname{tg} \delta\left(1+\frac{\sqrt{3}-3}{4}\right)=\frac{3+\sqrt{3}}{4} \quad \text { and } \\
\operatorname{tg} \delta=\frac{3+\sqrt{3}}{1+\sqrt{3}}=\frac{(3+\sqrt{3})(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}=\frac{3+\sqrt{3}-3 \sqrt{3}-3}{1-3}=\sqrt{3}
\end{gathered}
$$

Therefore, $\delta=A D B \varangle=60^{\circ}$.

II. solution. In the quadrilateral $A B D C$, the angles at the opposite vertices $A$ and $D$ are right angles, so their sum is $180^{\circ}$. This means that the quadrilateral is cyclic, and a circle can be inscribed around it.

![](https://cdn.mathpix.com/cropped/2024_05_02_7f8170a68b4295f75174g-1.jpg?height=441&width=442&top_left_y=1899&top_left_x=836)

$A C B \varangle=A D B \varangle$ are inscribed angles subtending the same arc, hence they are equal, which implies that $A D B \varangle=60^{\circ}$.