12.298. In a right-angled triangle, a plane is drawn through its hypotenuse, forming an angle $\alpha$ with the plane of the triangle, and an angle $\beta$ with one of the legs. Find the angle between this plane and the second leg.
Solution.

Through the hypotenuse $AB$ of the triangle $ABC$, a plane $\pi$ is drawn (Fig. 12.160).

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1052.jpg?height=434&width=638&top_left_y=756&top_left_x=628)

Fig. 12.160

$CD \perp \pi$. Then $\angle CAD$ and $\angle CBD$ are the angles between this plane and the legs of the triangle, $\angle CBD=\beta$.

Let $\angle CAD=\gamma, CD=1, CE$ be the altitude of $\triangle ABC$ drawn to its hypotenuse. Then its projection $DE$ on the plane $\pi$ is perpendicular to $AB$ and $\angle CED$ is the angle between the planes $ABC$ and $\pi, \angle CED=\alpha$.

In $\triangle ADC\left(\angle ADC=90^{\circ}\right): AC=\frac{CD}{\sin \gamma}=\frac{1}{\sin \gamma}$.

In $\triangle BDC\left(\angle BDC=90^{\circ}\right): BC=\frac{CD}{\sin \beta}=\frac{1}{\sin \beta}$.

In $\triangle CDE\left(\angle CDE=90^{\circ}\right): CE=\frac{CD}{\sin \alpha}=\frac{1}{\sin \alpha}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1053.jpg?height=693&width=767&top_left_y=17&top_left_x=256)

Fig. 12.161

In $\triangle ACB\left(\angle ACB=90^{\circ}\right)$:

$AB=\sqrt{AC^{2}+BC^{2}}=\sqrt{\frac{1}{\sin ^{2} \gamma}+\frac{1}{\sin ^{2} \beta}}=\frac{\sqrt{\sin ^{2} \beta+\sin ^{2} \gamma}}{\sin \gamma \sin \beta}$.

The area of triangle $ABC: S=\frac{1}{2} AC \cdot BC=\frac{1}{2} AB \cdot CE \Rightarrow$

$\Rightarrow \frac{1}{\sin \gamma \sin \beta}=\frac{\sqrt{\sin ^{2} \beta+\sin ^{2} \gamma}}{\sin \gamma \sin \beta \sin \alpha} \Leftrightarrow \sin \alpha=\sqrt{\sin ^{2} \beta+\sin ^{2} \gamma} \Rightarrow$

$\Rightarrow \sin \gamma=\sqrt{\sin ^{2} \alpha-\sin ^{2} \beta}=\sqrt{\frac{1-\cos 2 \alpha}{2}-\frac{1-\cos 2 \beta}{2}}=\sqrt{\frac{\cos 2 \beta-\cos 2 \alpha}{2}}=$ $=\sqrt{\sin (\alpha+\beta) \sin (\alpha-\beta)}$.

$\gamma=\arcsin \sqrt{\sin (\alpha+\beta) \sin (\alpha-\beta)}$.

Answer: $\arcsin \sqrt{\sin (\alpha+\beta) \sin (\alpha-\beta)}$.