## [ Two tangents drawn from one point

In triangle $A B C$, a circle is inscribed, touching side $A B$ at point $D$ and side $B C$ at point $E$. Find the angles of the triangle if $\frac{A D}{D B}=\frac{2}{1}$ and $\frac{B E}{E C}=\frac{1}{3}$.

#
Let $F$ be the point of tangency of the circle with side $A C$. Denote $B D=x$, then $A D=2 x$, and since the segments of tangents drawn from a point to a circle are equal, $B E=B D=x$ and $A F=A D=2 x$. Moreover, by the problem's condition, $E C=3 B E=3 x$, so $A B=A D+D B=2 x+x=3 x, B C=B E+E C=x+3 x=4 x, A C=A F+F C=2 x+3 x=5 x$.

$$
\begin{gathered}
\text { Then } \\
A C 2=25 \times 2=9 \times 2+16 \times 2=A B 2+B C 2, \\
\text { therefore, triangle } A B C \text { is a right triangle, } \\
\angle A B C=90^{\circ}, \cos \angle A C B=\frac{B C}{A C}=\frac{4 x}{5 x}=\frac{4}{5}, \sin \angle B A C=\cos \angle A C B=\frac{4}{5} .
\end{gathered}
$$

## Answer

$90^{\circ}, \arccos \frac{\mathbf{4}}{5}, \arcsin \frac{4}{5}$

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