B3. Let $\operatorname{tg} \alpha=\frac{a+2 b}{a \sqrt{3}}, \operatorname{tg} \beta=\frac{b+2 a}{b \sqrt{3}}, a>0, b>0, \alpha, \beta \in\left[0, \frac{\pi}{2}\right]$. Determine $\alpha+\beta$.
B3. Choose

$$
\operatorname{tg}(\alpha+\beta)=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \operatorname{tg} \beta}=\frac{\frac{a+2 b}{a \sqrt{3}}+\frac{b+2 a}{b \sqrt{3}}}{1-\frac{(a+2 b)(b+2 a)}{3 a b}}
$$

which can be simplified to $\mathrm{v} \operatorname{tg}(\alpha+\beta)=-\sqrt{3}$. The only possibility is $\alpha+\beta=\frac{2 \pi}{3}$.