1. In an acute-angled $\triangle A B C$, the angles at vertices $A, B$, and $C$ are denoted by $\alpha, \beta$, and $\gamma$, respectively. If

$$
4 \sin \alpha+5 \cos \beta=6
$$

and

$$
5 \sin \beta+4 \cos \alpha=5
$$

determine $\gamma$.
1. Let's square both equalities given in the problem and then add them. We obtain:

$$
\begin{aligned}
61 & =\left(16 \sin ^{2} \alpha+40 \sin \alpha \cos \beta+25 \cos ^{2} \beta\right)+\left(25 \sin ^{2} \beta+40 \sin \beta \cos \alpha+16 \cos ^{2} \alpha\right) \\
& =16\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)+25\left(\sin ^{2} \beta+\cos ^{2} \beta\right)+40(\sin \alpha \cos \beta+\sin \beta \cos \alpha) \\
& =16+25+40 \sin (\alpha+\beta)=41+40 \sin \left(180^{\circ}-\gamma\right)=41+40 \sin \gamma .
\end{aligned}
$$

From this, we have $\sin \gamma=\frac{61-41}{40}=\frac{20}{40}=\frac{1}{2}$, i.e., $\gamma=30^{\circ}$ or $\gamma=150^{\circ}$. The second possibility is ruled out due to the condition that $\triangle A B C$ is acute, so we are left with $\gamma=30^{\circ}$.