2B. For the angles $\alpha, \beta, \gamma$ of 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. According to the sine theorem, we have:

$$
\frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}=2 R
$$

so according to this,

$$
1=\frac{\sin ^{2} \beta+\sin ^{2} \gamma-\sin ^{2} \alpha}{\sin \beta \sin \gamma}=\frac{\frac{b^{2}}{4 R^{2}}+\frac{c^{2}}{4 R^{2}}-\frac{a^{2}}{4 R^{2}}}{\frac{b c}{4 R^{2}}}=\frac{b^{2}+c^{2}-a^{2}}{b c}
$$

Thus,

$$
a^{2}=b^{2}+c^{2}-b c
$$

From the cosine theorem, we have

$$
a^{2}=b^{2}+c^{2}-2 b c \cos \alpha
$$

From (1) and (2), we get $2 \cos \alpha=1$ and thus $\alpha=\frac{\pi}{3}$.

NOTE. If the vertices of a triangle lie on the hyperbola $x y=1$, then its orthocenter lies on that hyperbola. Prove it!