6. A circle of radius 2, inscribed in triangle $A B C$, touches side $B C$ at point $D$. A circle of radius 4 touches the extensions of sides $A B$ and $A C$, as well as side $B C$ at point $E$. Find the area of triangle $A B C$, if the measure of angle $A C B$ is $120^{\circ}$.
Solution. The centers of the circles lie on the angle bisectors. Therefore, \( C D = r / \operatorname{tg} 60^{\circ} = 2 / \sqrt{3}, \quad C E = R / \operatorname{tg} 30^{\circ} = 4 \sqrt{3}, D E = 4 \sqrt{3} - 2 / \sqrt{3} = 10 / \sqrt{3} \).

![](https://cdn.mathpix.com/cropped/2024_05_06_dbf1970871b608540faag-08.jpg?height=645&width=414&top_left_y=1188&top_left_x=181)

\[
\operatorname{tg}(D E N) = N D / E D = 4 / (10 / \sqrt{3}) = 2 \sqrt{3} / 5, \text { so }
\]

\[
\begin{aligned}
& \frac{A C}{\sin (D E N)} = \frac{E C}{\sin \left(60^{0} - D E N\right)} \Rightarrow A C = \frac{4 \sqrt{3} \sin (D E N)}{\sqrt{3} / 2 \cos D E N - 1 / 2 \sin D E N} \Rightarrow \\
& A C = \frac{4 \sqrt{3}}{\sqrt{3} / 2 \cdot 5 / (2 \sqrt{3}) - 1 / 2} = \frac{16}{\sqrt{3}} . \quad \text { The semiperimeter is } \\
& p = 16 / \sqrt{3} + 4 \sqrt{3}, \Rightarrow S = p r = 32 / \sqrt{3} + 8 \sqrt{3} = 56 / \sqrt{3}
\end{aligned}
\]

Answer: \( 56 / \sqrt{3} \)

## Solution for Variant № 10-7