For a triangle with angles $\alpha, \beta, \gamma$,

$$
\sin \alpha+\sin \beta=(\cos \alpha+\cos \beta) \sin \gamma
$$

is satisfied. What is the measure of angle $\gamma$?
I. solution. $\sin \gamma \neq 0$ (since it is about the angle of a triangle), so we can divide by $\sin \gamma$:

$$
\frac{\sin \alpha}{\sin \gamma}+\frac{\sin \beta}{\sin \gamma}=\cos \alpha+\cos \beta
$$

Based on the Law of Sines, we get:

$$
\frac{a}{c}+\frac{b}{c}=\cos \alpha+\cos \beta
$$

Write the Law of Cosines for $a$ and $b$, and express the values of $\cos \alpha$ and $\cos \beta$. Substituting these, we get:

$$
\frac{a+b}{c}=\frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{a^{2}+c^{2}-b^{2}}{2 a c}
$$

Multiply by $2 a b c$ and rearrange:

$$
\begin{gathered}
2 a b(a+b)=a b^{2}+a c^{2}-a^{3}+a^{2} b+b c^{2}-b^{3} \\
a^{3}+b^{3}+a b(a+b)=a c^{2}+b c^{2} \\
(a+b)\left(a^{2}-a b+b^{2}\right)+a b(a+b)=c^{2}(a+b)
\end{gathered}
$$

We can divide by $(a+b)$, and we get $a^{2}+b^{2}=c^{2}$.

According to the converse of the Pythagorean theorem, the triangle is right-angled: $\gamma=90^{\circ}$.

II. solution. Substitute $\left(180^{\circ}-\alpha-\beta\right)$ for $\gamma$:

$$
\begin{aligned}
\sin \alpha+\sin \beta & =(\cos \alpha+\cos \beta) \cdot \sin \left(180^{\circ}-\alpha-\beta\right)=(\cos \alpha+\cos \beta) \cdot \sin (\alpha+\beta)= \\
& =(\cos \alpha+\cos \beta)(\sin \alpha \cos \beta+\cos \alpha \sin \beta)= \\
& =\cos \alpha \cos \beta \sin \alpha+\cos \alpha \cos \beta \sin \beta+\cos ^{2} \alpha \sin \beta+\cos ^{2} \beta \sin \alpha
\end{aligned}
$$

This can also be rearranged into the following form:

$$
\begin{aligned}
& (\sin \alpha+\sin \beta)(1-\cos \alpha \cos \beta)=\cos ^{2} \alpha \sin \beta+\cos ^{2} \beta \sin \alpha \\
& (\sin \alpha+\sin \beta)(1-\cos \alpha \cos \beta)=\sin \beta-\sin ^{2} \alpha \sin \beta+\sin \alpha-\sin ^{2} \beta \sin \alpha \\
& (\sin \alpha+\sin \beta)(1-\cos \alpha \cos \beta)=(\sin \alpha+\sin \beta)(1-\sin \alpha \sin \beta)
\end{aligned}
$$

that is:

$$
(\sin \alpha+\sin \beta)(\sin \alpha \sin \beta-\cos \alpha \cos \beta)=0
$$

1. case: $\sin \alpha+\sin \beta=0$. This is not possible because $\alpha$ and $\beta$ are angles of a triangle.
2. case: $\sin \alpha \sin \beta=\cos \alpha \cos \beta$.

If $\cos \alpha$ or $\cos \beta$ were equal to 0, then at least one of $\sin \alpha$ and $\sin \beta$ would also have to be 0, but this is not possible.

If $\cos \alpha \cos \beta \neq 0$, then the equation can be written as: $\operatorname{tg} \alpha \operatorname{tg} \beta=1$.

Calculate $\operatorname{tg} \gamma$:

$$
\operatorname{tg} \gamma=\operatorname{tg}\left(180^{\circ}-\alpha-\beta\right)=-\operatorname{tg}(\alpha+\beta)=-\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \operatorname{tg} \beta}
$$

Since $\operatorname{tg} \alpha \operatorname{tg} \beta=1, \operatorname{tg} \gamma$ is undefined. This now means that $\gamma=90^{\circ}$, that is, the triangle is right-angled.