## Task B-3.3.

Let $\alpha$ and $\beta$ be the measures of two angles in a triangle whose circumradius is $6 \mathrm{~cm}$. Determine the sine of the third angle of this triangle and the length of the side opposite to it, if the following equalities hold:

$$
3 \sin \alpha + 4 \cos \beta = 6, \quad 4 \sin \beta + 3 \cos \alpha = 2
$$
## Solution.

Let's square the given equalities:

$$
\left\{\begin{array}{l}
9 \sin ^{2} \alpha+24 \sin \alpha \cos \beta+16 \cos ^{2} \beta=36 \\
16 \sin ^{2} \beta+24 \sin \beta \cos \alpha+9 \cos ^{2} \alpha=4
\end{array}\right.
$$

By adding these equalities, we get

$$
\begin{gathered}
9+24(\sin \alpha \cos \beta+\sin \beta \cos \alpha)+16=40 \\
24 \sin (\alpha+\beta)=15 \\
\sin (\alpha+\beta)=\frac{5}{8}
\end{gathered}
$$

Since the third angle of the given triangle is $\gamma=180^{\circ}-(\alpha+\beta)$, it follows that $\sin \gamma=\sin (\alpha+\beta)$.

Therefore, $\sin \gamma=\frac{5}{8}$.

The length of the opposite side is $c=2 R \sin \gamma=2 \cdot 6 \cdot \frac{5}{8}=7.5 \mathrm{~cm}$.