[ Inscribed and Circumscribed Circles [ $\underline{\text { T }}$ rigonometric Ratios in a Right Triangle ]

In triangle $A B C$ with perimeter $2 p$, the measure of the acute angle $A B C$ is $\alpha$ and $A C=a$. A circle is inscribed in the triangle with its center at point $O$. Find the area of triangle $A O C$.

#
The distance from the vertex of a triangle to the nearest point of tangency with the inscribed circle is equal to the difference between the semiperimeter and the opposite side of the triangle.

## Solution

Let $K$ be the point of tangency of the inscribed circle with side $BC$. Then,

$$
B K = p - a, \quad O K = B K \operatorname{tg} \frac{\alpha}{2} = (p - a) \operatorname{tg} \frac{\alpha}{2}
$$

Since $O K$ is the radius of the inscribed circle, then

$$
S_{\triangle \mathrm{AOC}} = \frac{1}{2} a(p - a) \operatorname{tg} \frac{\alpha}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-11.jpg?height=449&width=869&top_left_y=2445&top_left_x=595)

## Problem