1B. Find the angle $\alpha$ of triangle $ABC$, if for its angles $\alpha, \beta, \gamma$ the equality $\sin \alpha=\frac{\sin \beta+\sin \gamma}{\cos \beta+\cos \gamma}$ holds.
Solution. Since $\alpha, \beta, \gamma$ are angles of a triangle, from $\sin \alpha=\frac{\sin \beta+\sin \gamma}{\cos \beta+\cos \gamma}$ we get

$$
\begin{aligned}
& \cos \beta+\cos \gamma \neq 0, \cos \frac{\beta-\gamma}{2} \neq 0, \frac{\beta+\gamma}{2}=\frac{\pi}{2}-\frac{\alpha}{2} \text { so } \\
& \sin \alpha=\frac{\sin \beta+\sin \gamma}{\cos \beta+\cos \gamma}=\frac{2 \sin \frac{\beta+\gamma}{2} \cos \frac{\beta-\gamma}{2}}{2 \cos \frac{\beta+\gamma}{2} \cos \frac{\beta-\gamma}{2}} \Rightarrow 2 \sin ^{2} \frac{\alpha}{2} \cos \frac{\alpha}{2}=\cos \frac{\alpha}{2} \quad \Rightarrow \\
& \cos \frac{\alpha}{2}=0 \text { or } \sin ^{2} \frac{\alpha}{2}=1 \quad \Rightarrow \alpha=\pi \text { which is not possible, or } \alpha=\frac{\pi}{2} \text {. }
\end{aligned}
$$