10.013. Given a triangle with sides 12, 15, and \(18 \, \text{cm}\). A circle is drawn that touches both smaller sides and has its center on the larger side. Find the segments into which the center of the circle divides the larger side of the triangle.
## Solution.

Using the cosine theorem, from Fig. 10.13 we have:

$$
\begin{aligned}
& A B^{2}=B C^{2}+A C^{2}-2 \cdot B C \cdot A C \cdot \cos \alpha \Rightarrow \\
& \Rightarrow 12^{2}=15^{2}+18^{2}-2 \cdot 15 \cdot 18 \cos \alpha, \cos \alpha=\frac{3}{4}, \sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\frac{\sqrt{7}}{4} \\
& B C^{2}=A B^{2}+A C^{2}-2 \cdot A B \cdot A C \cdot \cos \beta \Rightarrow \\
& \Rightarrow 15^{2}=12^{2}+18^{2}-2 \cdot 12 \cdot 18 \cdot \cos \beta, \cos \beta=\frac{9}{16}, \sin \beta=\sqrt{1-\cos ^{2} \beta}=\frac{5 \sqrt{7}}{16}
\end{aligned}
$$

$$
\begin{aligned}
& \sin \alpha=\frac{O M}{O C}=\frac{R}{18-x}, \sin \beta=\frac{O K}{A O}=\frac{R}{x} \text { and } \frac{\sin \alpha}{\sin \beta}=\frac{x}{18-x}=\frac{4}{5} \\
& 5 x=72-4 x, 9 x=72, x=8, y=10 . A O=8 \text{ cm}, O C=10 \text{ cm}
\end{aligned}
$$

Answer: 8 and 10 cm.