6. (5 points) From point $A$, lying on a circle of radius 3, chords $A B, A C$ and tangent $A D$ are drawn. The angle between the chords is $\frac{\pi}{4}$, and the angle between chord $A C$ and tangent $A D$, which does not contain chord $A B$, is $\frac{5 \pi}{12}$. Calculate the integer area of triangle $A B C$.

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# Solution:

Let $\angle D A C=\alpha, \angle B A C=\beta$, and the radius of the circle be $R$. It is known that $\angle A C B=\angle D A C=\alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f859f267290c335bb97ag-06.jpg?height=583&width=599&top_left_y=271&top_left_x=320)

By the Law of Sines, $\quad \frac{|A B|}{\sin \alpha}=\frac{|B C|}{\sin \beta}=2 R$. Therefore, $\quad|A B|=2 R \sin \alpha$, and $|B C|=2 R \sin \beta$.

$S_{\triangle A B C}=\frac{1}{2}|A B| \cdot|B C| \cdot \sin (\angle A B C)$.

Thus,

$S_{\triangle A B C}=2 R^{2} \sin \alpha \sin \beta \sin (\alpha+\beta)$

To calculate the area, we need to know the values of $\sin \alpha$ and $\sin (\alpha+\beta)$. We will compute these values under the conditions of the problem.

For this, we will calculate $\sin \alpha$ and $\cos \alpha$.

$$
\begin{aligned}
& \sin \alpha=\sin \frac{5 \pi}{12}=\sin \left(\frac{\pi}{2}-\frac{\pi}{12}\right)=\cos \frac{\pi}{12}=\sqrt{\frac{1}{2}\left(1+\cos \frac{\pi}{6}\right)}=\frac{1}{2} \sqrt{2+\sqrt{3}}=\frac{1+\sqrt{3}}{2 \sqrt{2}} \\
& \cos \alpha=\cos \frac{5 \pi}{12}=\cos \left(\frac{\pi}{2}-\frac{\pi}{12}\right)=\sin \frac{\pi}{12}=\sqrt{\frac{1}{2}\left(1-\cos \frac{\pi}{6}\right)}=\frac{1}{2} \sqrt{2-\sqrt{3}}=\frac{-1+\sqrt{3}}{2 \sqrt{2}} \\
& \sin (\alpha+\beta)=\sin \left(\frac{\pi}{4}+\frac{5 \pi}{12}\right)=\sin \frac{2 \pi}{3}=\frac{\sqrt{3}}{2}
\end{aligned}
$$

Substituting the obtained values into (2), it is not difficult to get that $S_{\triangle A B C}=\frac{9}{4}(3+\sqrt{3})$.

It remains to compute the answer.

Answer: 10.