## Task B-3.3.

Within the angle $\varangle p O q$ measuring $60^{\circ}$, a point $A$ is chosen such that it is $1 \mathrm{~cm}$ away from one arm and $2 \mathrm{~cm}$ away from the other arm. Calculate the length of the line segment $\overline{O A}$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_bcbf631b6726be7fa637g-19.jpg?height=680&width=694&top_left_y=491&top_left_x=630)

With the labels as in the figure, we have

$$
\sin \alpha=\frac{1}{x} \quad \text{and} \quad \sin \left(60^{\circ}-\alpha\right)=\frac{2}{x}
$$

By dividing these equations, we get:

$$
\begin{gathered}
\frac{\sin \alpha}{\sin 60^{\circ} \cos \alpha - \cos 60^{\circ} \sin \alpha} = \frac{1}{2} \\
\frac{\sin \alpha}{\frac{\sqrt{3}}{2} \cos \alpha - \frac{1}{2} \sin \alpha} = \frac{1}{2} \\
2 \sin \alpha = \frac{\sqrt{3}}{2} \cos \alpha - \frac{1}{2} \sin \alpha \\
5 \sin \alpha = \sqrt{3} \cos \alpha \\
\tan \alpha = \frac{\sqrt{3}}{5}
\end{gathered}
$$

Further, since $1 + \cot^2 \alpha = \frac{1}{\sin^2 \alpha}$, we have $\frac{1}{\sin^2 \alpha} = 1 + \frac{25}{3} = \frac{28}{3}$.

Since $\alpha$ is an acute angle, then $\sin \alpha = +\sqrt{\frac{3}{28}}$.

Finally, from $\sin \alpha = \frac{1}{x}$, we get $x = \sqrt{\frac{28}{3}} = \frac{2 \sqrt{21}}{3} \text{ cm}$.