12.299. In a right-angled triangle with an acute angle $\alpha$, a plane is drawn through the smallest median, forming an angle $\beta$ with the plane of the triangle. Find the angles between this plane and the legs of the triangle.
## Solution.

Let the median $D E$, drawn to the larger side of the triangle

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1054.jpg?height=747&width=802&top_left_y=77&top_left_x=269)

Fig. 12.162

$A D F$ - its hypotenuse $A F$ - the smaller median of the given triangle (Fig. 12.161), through which the plane $\pi, \angle A F D=\alpha$ passes.

From point $A$, draw perpendiculars $A B$ and $A C$ to the plane $\pi$ and to $D E$. Then $B C$ is the projection of $A C$ onto the plane $\pi, D B$ is the projection of $D A$ onto the plane $\pi, B C \perp D E, \quad \angle A C B$ is the angle between the planes $A D F$ and $\pi$, $\angle A C B=\beta, \angle A D B$ is the angle between the leg $A D$ and the plane $\pi$.

Let $\angle A D B=\gamma, A C=1$.

Since $D E=E F$, then $\angle C D F=\angle D F E=\alpha \Rightarrow \angle A D E=90^{\circ}-\alpha, \angle D A C=\alpha$.

In $\triangle A B C\left(\angle A B C=90^{\circ}\right): A B=A C \sin \angle A C B=\sin \beta$.

In $\triangle A C D\left(\angle A C D=90^{\circ}\right): A D=\frac{A C}{\cos \angle D A C}=\frac{1}{\cos \alpha}$.

In $\triangle A B D\left(\angle A B D=90^{\circ}\right): \sin \gamma=\frac{A B}{A D}=\sin \beta \cos \alpha, \gamma=\arcsin (\sin \beta \cos \alpha)$.

To find the angle between the plane $\pi$ and the second leg, replace the angle $\alpha$ with the angle $90^{\circ}-\alpha$, and we get that it is equal to $\arcsin (\sin \beta \sin \alpha)$.

Answer: $\arcsin (\sin \beta \cos \alpha) \arcsin (\sin \beta \sin \alpha)$.