3. Find the side $B C$ of the quadrilateral $A B C D$, if $\angle B A C=\alpha, \angle A C D=\beta, \angle B C A+\angle C A D=\frac{\pi}{2}$ and $A D=a$. In the answer, write the result rounded to two decimal places.

$$
\alpha=\arcsin \frac{5}{13}, \beta=\arcsin \frac{12}{13}, a=24
$$
Solution. Since the sum of angles $\angle B A C+\angle A C D=\arcsin \frac{5}{13}+\arcsin \frac{12}{13}=\frac{\pi}{2}$, then $\angle B A D+\angle B C D=$ $\pi$. Therefore, a circle can be circumscribed around quadrilateral $A B C D$. Next, by the Law of Sines $\frac{A D}{\sin \beta}=\frac{B C}{\sin \alpha} \Rightarrow B C=\frac{24 \cdot 5}{12}=10$.

Answer: 10.