## Solution.

Let $K, L$, and $M$ be the feet of the perpendiculars from point $T$ to the sides $\overline{B C}, \overline{C A}$, and $\overline{A B}$, respectively, and let $d$ be the length of the side of the equilateral triangle $K L M$. Denote the angles and side lengths of triangle $A B C$ in the usual way as $\alpha, \beta$, and $\gamma$, and $a, b$, and $c$.

![](https://cdn.mathpix.com/cropped/2024_05_30_a630fb9dab34a5225850g-17.jpg?height=813&width=1071&top_left_y=453&top_left_x=421)

Notice that the quadrilateral $A M T L$ is cyclic because it has two opposite right angles.

The length $\overline{A T}$ is the diameter of the circumcircle of this quadrilateral, so we have

$$
d=|L M|=|A T| \sin \alpha=56 \sin \alpha
$$

Similarly, we show that $d=40 \sin \beta$ and $d=35 \sin \gamma$.

Using the Law of Sines for triangle $A B C$, we conclude

$$
\frac{a}{b}=\frac{\sin \alpha}{\sin \beta}=\frac{40}{56}=\frac{5}{7}
$$

and similarly, $\frac{c}{b}=\frac{40}{35}=\frac{8}{7}$.

By the Law of Cosines for triangle $A B C$, we have $b^{2}=c^{2}+a^{2}-2 c a \cos \beta$, so dividing by $b^{2}$ gives

$$
1=\left(\frac{c}{b}\right)^{2}+\left(\frac{a}{b}\right)^{2}-2 \cdot \frac{c}{b} \cdot \frac{a}{b} \cdot \cos \beta
$$

Substituting the obtained ratios, we get

$$
1=\frac{64}{49}+\frac{25}{49}-\frac{80}{49} \cos \beta
$$

Simplifying, we find $\cos \beta=\frac{1}{2}$, so $\beta=60^{\circ}$.
## Task A-3.4.

Inside the triangle $ABC$ there is a point $T$ such that $|AT|=56,|BT|=40,|CT|=$ 35. The feet of the perpendiculars from point $T$ to the sides of triangle $ABC$ are the vertices of an equilateral triangle. Determine the angle $\varangle ABC$.