2. In the right triangle $A B C$, the leg $\overline{C A}=b$ and the angle $\alpha=60^{\circ}$ at vertex $A$ are given. From vertex $C$, a perpendicular $C D$ is dropped to the hypotenuse $A B$. From point $D$, a perpendicular $D C_{1}$ is dropped to the leg $B C$. From point $C_{1}$, a perpendicular $C_{1} D_{1}$ is dropped to the hypotenuse $A B$, etc. Calculate the sum of the lengths of the thus obtained perpendiculars when their number increases indefinitely.
Solution. The angles $\measuredangle C A D, \measuredangle D C C_{1}, \measuredangle C_{1} D D_{1}, \ldots$ are equal to each other and are $60^{\circ}$, because they are angles with mutually perpendicular sides (see diagram). From the right-angled triangles $A C D, D C C_{1}, D D_{1} C_{1}, \ldots$ we get

$$
\begin{array}{ll}
\overline{C D}=b \sin 60^{\circ}, & \overline{C_{1} D}=\overline{C D} \cdot \sin 60^{\circ} \\
\overline{C_{1} D_{1}}=\overline{C_{1} D} \cdot \sin 60^{\circ}, \ldots & \overline{C D}=b \sin 60^{\circ} \\
\overline{C_{1} D}=b \cdot \sin ^{2} 60^{\circ}, & \overline{C_{1} D_{1}}=b \cdot \sin ^{3} 60^{\circ}
\end{array}
$$

$$
\ldots
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_e6c03bcb70f03208de85g-6.jpg?height=300&width=548&top_left_y=1883&top_left_x=916)

The sum of these segments (when their number increases indefinitely) forms an infinite geometric series with the first term $a_{1}=b \sin 60^{\circ}$ and common ratio

$$
q=\sin 60^{\circ}=\frac{\sqrt{3}}{2}<1
$$

According to this, the sum $S$ of this series will be

$$
S=\frac{a_{1}}{1-q}=\frac{b \cdot \sin 60^{\circ}}{1-\sin 60^{\circ}}=b \sqrt{3}(2+\sqrt{3})
$$