$\left[\begin{array}{l}\text { Inscribed and Circumscribed Circles } \\ {[\quad \text { Law of Sines }}\end{array}\right]$

In an isosceles triangle, the base is 24, and the lateral side is 15. Find the radii of the inscribed and circumscribed circles.
To find the radius of the inscribed circle, we can use the property of the angle bisector of a triangle. For the radius of the circumscribed circle, we use the formula $a=2 R \sin \boldsymbol{\alpha}$.

## Solution

Let $A M$ be the height of the isosceles triangle $A B C$, where $A B=A C=15$, $B C=24$, and $O$ is the center of the inscribed circle, with $r$ and $R$ being the radii of the inscribed and circumscribed circles, respectively.

Since the triangle is isosceles, the point $O$ lies on the segment $A M$, and $B O$ is the angle bisector of $\angle A B C$.

Therefore, $\frac{O M}{O A}=\frac{B M}{A B}$. Consequently,

$$
\begin{gathered}
r=O M=\frac{A M \cdot M B}{M B+A B}=\frac{9 \cdot 12}{12+15}=4 \\
R=\frac{A C}{2 \sin \angle A B C}=\frac{A C}{2 \cdot \frac{A M}{A B}}=\frac{15}{2 \cdot \frac{9}{15}}=\frac{25}{2}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-17.jpg?height=434&width=694&top_left_y=0&top_left_x=682)

## Answer

$4 ; 12.5$.