Apply the Law of Cosines.

## Solution

By the Law of Cosines in triangle $B K C$, we find that

$$
B K=\sqrt{C K^{2}+C B^{2}-2 C K \cdot C B \cos \angle K C B}=\sqrt{9+25-2 \cdot 3 \cdot 5 \cdot \frac{4}{5}}=\sqrt{10}.
$$

If $R$ is the radius of the circumcircle of triangle $B K C$, then

$$
R=\frac{B K}{2 \sin \angle K C B}=\frac{\sqrt{10}}{2 \cdot \frac{3}{5}}=\frac{5 \sqrt{10}}{6}
$$

By the Law of Cosines in triangle $A B C$, we find that

$$
A B=\sqrt{A C^{2}+B C^{2}-2 A C \cdot B C \cos \angle A C B}=\sqrt{16+25-2 \cdot 4 \cdot 5 \cdot \frac{4}{5}}=3
$$

Therefore, triangle $B A K$ is a right triangle ( $B K^{2}=10=1+9=A K^{2}+A B^{2}$ ). Thus, the radius of the inscribed circle in this triangle can be calculated by the formula

$$
r=\frac{A B+A K-B K}{2}, \text{ i.e., } r=\frac{4-\sqrt{10}}{2}
$$

Therefore,

$$
\frac{R}{r}=\frac{5 \sqrt{10}}{(3(4-\sqrt{10})}=\frac{10 \sqrt{10}+25}{9}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-06.jpg?height=397&width=598&top_left_y=1248&top_left_x=730)

## Answer

$\frac{25+10 \sqrt{10}}{9}$ Send a comment

## Problem
In triangle $A B C$, side $B C$ is equal to 5. A circle passes through vertices $B$ and $C$ and intersects side $A C$ at point $K$, such that $C K=3, K A=1$. It is known that the cosine of angle $A C B$ is $\frac{4}{5}$. Find the ratio of the radius of the given circle to the radius of the circle inscribed in triangle $A B K$.