## Task A-3.6.

Let $A B C$ be a triangle in which $\operatorname{tg} \varangle B A C=1$ and $\operatorname{tg} \varangle A B C=2$.

Determine the ratio $|B C|:|A B|$.
## First solution.

Draw the altitude from vertex $C$ to side $\overline{A B}$ and denote its foot by $N$.

Angles $\varangle B A C$ and $\varangle A B C$ are acute (since their tangents are positive), so from the right triangles $C A N$ and $C B N$ we see that

$$
\begin{aligned}
& 1=\operatorname{tg} \varangle B A C=\operatorname{tg} \varangle N A C=\frac{|C N|}{|A N|}, \\
& 2=\operatorname{tg} \varangle A B C=\operatorname{tg} \varangle N B C=\frac{|C N|}{|B N|} .
\end{aligned}
$$

It follows that

$$
|B N|:|A N|=1: 2
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_5bba13a74780d91c8bdbg-21.jpg?height=512&width=699&top_left_y=178&top_left_x=610)

Let $|B N|=x$ and $|A N|=|C N|=2 x$. Then

$$
|A B|=|A N|+|N B|=2 x+x=3 x
$$

and applying the Pythagorean theorem to the right triangle $C B N$ we get

$$
|B C|=\sqrt{|C N|^{2}+|B N|^{2}}=\sqrt{4 x^{2}+x^{2}}=\sqrt{5} x
$$

We conclude that

$$
|B C|:|A B|=\sqrt{5} x: 3 x=\sqrt{5}: 3
$$