6. Two acute angles $\alpha$ and $\beta$ satisfy the condition $\operatorname{Sin}^{2} \alpha+\operatorname{Sin}^{2} \beta=\operatorname{Sin}(\alpha+\beta)$. Find the sum of the angles $\alpha+\beta$ in degrees.
6. Using the formula for $\operatorname{Sin}(\alpha+\beta)$, we get:

$\operatorname{Sin}^{2} \alpha+\operatorname{Sin}^{2} \beta=\operatorname{Sin} \alpha \operatorname{Cos} \beta+\operatorname{Cos} \alpha \operatorname{Sin} \beta \Leftrightarrow \operatorname{Sin} \alpha(\operatorname{Sin} \alpha-\operatorname{Cos} \beta)=\operatorname{Sin} \beta(\operatorname{Cos} \alpha-\operatorname{Sin} \beta)$. Since $\alpha, \beta$ are acute angles, the brackets $\operatorname{Sin} \alpha-\operatorname{Cos} \beta$ and $\operatorname{Cos} \alpha-\operatorname{Sin} \beta$ have the same sign. Let them be positive, then:

$\left\{\begin{array}{l}\operatorname{Sin} \alpha-\operatorname{Cos} \beta>0 \\ \operatorname{Cos} \alpha-\operatorname{Sin} \beta>0\end{array} \Leftrightarrow \Leftrightarrow\left\{\begin{array}{l}\operatorname{Sin} \alpha>\operatorname{Cos} \beta \\ \operatorname{Cos} \alpha>\operatorname{Sin} \beta\end{array} \Leftrightarrow\left\{\begin{array}{l}\operatorname{Sin}^{2} \alpha>\operatorname{Cos}^{2} \beta \\ \operatorname{Cos}^{2} \alpha>\operatorname{Sin}^{2} \beta\end{array}\right.\right.\right.$ $\Rightarrow \operatorname{Sin}^{2} \alpha+\operatorname{Cos}^{2} \alpha>\operatorname{Cos}^{2} \beta+\operatorname{Sin}^{2} \beta=1$, which is impossible. Similarly, it is impossible that $\operatorname{Sin}^{2} \beta+\operatorname{Cos}^{2} \beta<1$. Therefore, $\operatorname{Sin} \alpha=\operatorname{Cos} \beta, \operatorname{Cos} \alpha=\operatorname{Sin} \beta \Rightarrow \operatorname{Sin} \alpha=\operatorname{Sin}(90-\beta) \Rightarrow \alpha+\beta=90^{\circ}$. A similar proof applies if the brackets $\operatorname{Sin} \alpha-\operatorname{Cos} \beta$ and $\operatorname{Cos} \alpha-\operatorname{Sin} \beta$ are both negative. Answer: $\left\{90^{\circ}\right\}$