The tower in King Arthur's castle is topped with a roof that represents a triangular pyramid, where all the planar angles at the vertex are right angles. The three slopes of the roof are painted in different colors. The red slope of the roof is inclined to the horizontal at an angle $\alpha$, and the blue slope - at an angle $\beta$. Find the probability that a raindrop, falling vertically onto the roof at a random place, lands on the green slope.

![](https://cdn.mathpix.com/cropped/2024_05_08_22585010b2534c7ab9ebg-42.jpg?height=589&width=515&top_left_y=0&top_left_x=766)
It is natural to assume that a drop falls on the roof randomly in the sense that the probability of it landing in a certain area of the roof is proportional to the area of the projection of this area onto a horizontal surface. Then, we should be interested in the areas of the projections of the slopes onto the triangular base of the roof.

![](https://cdn.mathpix.com/cropped/2024_05_08_22585010b2534c7ab9ebg-42.jpg?height=549&width=669&top_left_y=889&top_left_x=686)

Let the area of the green slope be $S_{g}$. Then the area $S_{O R B}$ of the projection of this slope onto the plane of the triangle $R B G$, lying at the base of the roof, is $S_{g} \cos \gamma$, where $\gamma$ is the unknown angle of inclination of the blue slope.

Let $S = S_{R G B}$. From the condition, it follows that each edge of the pyramid is perpendicular to the lateral face in which it does not lie. Therefore, $S_{g} = S \cos \gamma$. Consequently, $S_{O R B} = S \cos^2 \gamma$.

The desired probability is $\frac{S_{O R B}}{S} = \frac{S \cos^2 \gamma}{S} = \cos^2 \gamma$.

It remains to find $\cos^2 \gamma$. By performing similar calculations, we find: $S_{O B G} = S \cos^2 \alpha$, $S_{O R G} = S \cos^2 \beta$. Therefore, $S = S (\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma)$, from which

$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$. Hence, $\cos^2 \gamma = 1 - \cos^2 \beta - \cos^2 \alpha$.

## Answer

$1 - \sin^2 \beta - \cos^2 \alpha$.