3A. For the angles $\alpha, \beta$ and $\gamma$ in triangle $ABC$, the following equality holds:

$$
\frac{\sin ^{2} \beta+\sin ^{2} \gamma-\sin ^{2} \alpha}{\sin \beta \sin \gamma}=1
$$

Calculate the angle $\alpha$.
Solution. Applying the sine and cosine theorems, we transform the given equality as follows:

$$
1=\frac{\frac{b^{2}}{4 R^{2}}+\frac{c^{2}}{4 R^{2}}-\frac{a^{2}}{4 R^{2}}}{\frac{b}{2 R} \cdot \frac{c}{2 R}}=\frac{-a^{2}+b^{2}+c^{2}}{b c}=\frac{2 b c \cos \alpha}{b c}=2 \cos \alpha
$$

Therefore, $\cos \alpha=\frac{1}{2}$. Since $\alpha$ is an angle in a triangle, we get $\alpha=\frac{\pi}{3}$.