3.169. Find the number $\beta\left(\frac{\pi}{2}<\beta<\pi\right)$, if it is known that $\operatorname{tg}(\alpha+\beta)=\frac{9}{19}$ and $\operatorname{tg} \alpha=-4$.
Solution.

$$
\operatorname{tg}(\alpha+\beta)=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \operatorname{tg} \beta}=\frac{9}{19} .
$$

Since $\operatorname{tg} \alpha=-4$, then $\frac{-4+\operatorname{tg} \beta}{1+4 \operatorname{tg} \beta}=\frac{9}{19}, \operatorname{tg} \beta=-4$, from which
$\beta=-\operatorname{arctg} 5+\pi n, n \in Z$. Considering that $\beta \in\left(\frac{\pi}{2}, \pi\right)$, we get $\beta=\pi-\operatorname{arctg} 5$.

Answer: $\beta=\pi-\operatorname{arctg} 5$.