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Pythagorean Theorem (direct and inverse). [ Trigonometric ratios in a right triangle ]

The medians of a right triangle, drawn from the vertices of the acute angles, are in the ratio $\sqrt{2}: 1$. Find the acute angles of the triangle.

#
Express the specified medians in terms of the legs of the triangle.

## Solution

Let the legs of the triangle be $2 x$ and $2 y$. Then the squares of the specified medians are $4 x^{2}+y^{2}$ and $x^{2}+4 y^{2}$. From the condition of the problem, we get the equation

$$
\frac{4 x^{2}+y^{2}}{x^{2}+4 y^{2}}=2, \text{ or } 7 y^{2}=2 x^{2}
$$

Therefore, the tangent of one of the acute angles of the triangle is

$$
\frac{2 y}{2 x}=\frac{y}{x}=\sqrt{\frac{2}{7}}
$$

## Answer

$\operatorname{arctg} \sqrt{\frac{2}{7}}, \operatorname{arcctg} \sqrt{\frac{2}{7}}$.