Task B-3.8. For the angles $\alpha$ and $\beta$ of triangle $ABC$, the following holds:

$$
\begin{aligned}
& 3 \sin \alpha + 4 \cos \beta = 6 \\
& 4 \sin \beta + 3 \cos \alpha = 1
\end{aligned}
$$

Determine the measure of angle $\gamma$ of this triangle.
Solution. After squaring the given equalities, we get:

$$
\begin{aligned}
& 9 \sin ^{2} \alpha+24 \sin \alpha \cos \beta+16 \cos ^{2} \beta=36 \\
& 16 \sin ^{2} \beta+24 \sin \beta \cos \alpha+9 \cos ^{2} \alpha=1
\end{aligned}
$$

Adding these equalities, we obtain

$$
\begin{aligned}
& 9\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)+24(\sin \alpha \cos \beta+\sin \beta \cos \alpha)+16\left(\sin ^{2} \beta+\cos ^{2} \beta\right)=37 \\
& 24 \sin (\alpha+\beta)=12 \\
& \sin (\alpha+\beta)=\frac{1}{2}
\end{aligned}
$$

It follows that $\alpha+\beta=30^{\circ}$ or $\alpha+\beta=150^{\circ}$.

If $\alpha+\beta=30^{\circ}$, then $\alpha<30^{\circ}$ and $\sin \alpha<\frac{1}{2}$, which is in contradiction with the first equality. Indeed,

$$
3 \sin \alpha+4 \cos \beta<3 \cdot \frac{1}{2}+4<6
$$

so the first equality would not be satisfied.

Therefore, $\alpha+\beta=150^{\circ}$, and thus $\gamma=30^{\circ}$

(2 points)

## SCHOOL/CITY COMPETITION IN MATHEMATICS

## 4th grade - high school - B variant

24. January 2011.

IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO EVALUATE THIS APPROACH AND SCORE IT APPROPRIATELY.