4A. To find the tangents of the angles in a triangle, if they are integers.




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Solution. Let $\alpha, \beta$ and $\gamma$ be the angles of the triangle, for which, without loss of generality, we can assume that $\alpha<\beta<\gamma$. Then $\alpha \in\left(0, \frac{\pi}{3}\right]$, and therefore $\operatorname{tg} \alpha \in(0, \sqrt{3}]$.

Since $\operatorname{tg} \alpha$ is an integer, we get that $\operatorname{tg} \alpha=1$. Then

$$
-1=-\operatorname{tg} \alpha=\operatorname{tg}(\pi-\alpha)=\operatorname{tg}(\beta+\gamma)=\frac{\operatorname{tg} \beta+\operatorname{tg} \gamma}{1-\operatorname{tg} \beta \operatorname{tg} \gamma}
$$

From the last equality, we have

$$
\operatorname{tg} \beta+\operatorname{tg} \gamma=1-\operatorname{tg} \beta \operatorname{tg} \gamma
$$

which can be written in the form

$$
(\operatorname{tg} \beta-1)(\operatorname{tg} \gamma-1)=2
$$

Now it is clear that $\operatorname{tg} \beta-1=1$ and $\operatorname{tg} \gamma-1=2$. Therefore, the required tangents are

$$
\operatorname{tg} \alpha=1, \operatorname{tg} \beta=2 \text { and } \operatorname{tg} \gamma=3 .
$$