## Task B-2.3.

Let $\alpha, \beta$ and $\gamma$ be the angles of a triangle. If $\frac{\sin ^{2} \gamma+\sin ^{2} \beta-\sin ^{2} \alpha}{\sin \beta \cdot \sin \gamma}=\sqrt{3}$, determine the angle $\alpha$.
## Solution.

Let $a$ be the length of the side opposite to angle $\alpha$, $b$ be the length of the side opposite to angle $\beta$, and $c$ be the length of the side opposite to angle $\gamma$ of the given triangle.

By applying the Law of Sines, we have:

$$
\begin{aligned}
\frac{a}{\sin \alpha} & =2 R \Longrightarrow \sin \alpha=\frac{a}{2 R} \\
\frac{b}{\sin \beta} & =2 R \Longrightarrow \sin \beta=\frac{b}{2 R} \\
\frac{c}{\sin \gamma} & =2 R \Longrightarrow \sin \gamma=\frac{c}{2 R}
\end{aligned}
$$

Substituting the expressions for the sines into the given equation, we get

$$
\frac{\frac{c^{2}}{4 R^{2}}+\frac{b^{2}}{4 R^{2}}-\frac{a^{2}}{4 R^{2}}}{\frac{b}{2 R} \cdot \frac{c}{2 R}}=\sqrt{3}
$$

which simplifies to

$$
\frac{c^{2}+b^{2}-a^{2}}{b c}=\sqrt{3}
$$

or $c^{2}+b^{2}-a^{2}=\sqrt{3} b c$.

According to the Law of Cosines, $\cos \alpha=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$, so

$$
\cos \alpha=\frac{\sqrt{3} b c}{2 b c}=\frac{\sqrt{3}}{2}
$$

Thus, the desired angle $\alpha=30^{\circ}$.