12.428 Given three pairwise mutually perpendicular rays $O M, O N$ and $OP$. On the ray $O M$, a point $A$ is taken at a distance $O A$, equal to $a$; on the rays $O N$ and $O P$, points $B$ and $C$ are taken, respectively, such that the angle $A B C$ is equal to $\alpha$, and the angle $A C B$ is equal to $\beta$. Find $O B$ and $O C$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-304.jpg?height=286&width=324&top_left_y=1389&top_left_x=130)

Let $x = OB, y = OC$. Then

$$
AB^2 = a^2 + x^2, \quad AC^2 = a^2 + y^2, \quad BC^2 = x^2 + y^2
$$

By the Law of Sines for $\triangle ABC$, we have:

$$
\begin{aligned}
& \frac{a^2 + y^2}{\sin^2 \alpha} = \frac{a^2 + x^2}{\sin^2 \beta} = \frac{x^2 + y^2}{\sin^2 (\alpha + \beta)} \Rightarrow \\
& \Rightarrow \frac{a^2 + y^2}{\sin^2 \alpha} = \frac{a^2 - y^2}{\sin^2 \beta - \sin^2 (\alpha + \beta)} \Rightarrow
\end{aligned}
$$

$$
\Rightarrow \frac{a^2 + y^2}{\sin \alpha} = \frac{a^2 - y^2}{\sin (\alpha + 2 \beta)} \Rightarrow y^2 = a^2 \cdot \frac{\sin \alpha + \sin (\alpha + 2 \beta)}{\sin \alpha - \sin (\alpha + 2 \beta)} = a^2 (-\cot \beta) \cdot \tan (\alpha + \beta)
$$

Similarly, $x^2 = a^2 (-\cot \alpha) \cdot \tan (\alpha + \beta)$.

Answer:

$$
a \sqrt{-\cot \beta \cdot \tan (\alpha + \beta)}, \quad a \sqrt{-\cot \alpha \cdot \tan (\alpha + \beta)}
$$