Two angles of a triangle are $40^{\circ}$ and $80^{\circ}$. Find the angles of the triangle with vertices at the points of tangency of the inscribed circle with the sides of the given triangle.

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Let the angles $A$ and $B$ of triangle $ABC$ be $40^{\circ}$ and $80^{\circ}$, respectively. Then $\angle C=60^{\circ}$.

Denote by $A_{1}, B_{1}$, and $C_{1}$ the points of tangency of the incircle of triangle $ABC$ with the sides $BC, AC$, and $AB$, respectively. Then

$\angle A C_{1} B_{1}=1 / 2\left(180^{\circ}-40^{\circ}\right)=70^{\circ}$, and $\angle B C_{1} A_{1}=1 / 2\left(180^{\circ}-40^{\circ}\right)=50^{\circ}$.

Therefore, $\angle A_{1} C_{1} B_{1}=180^{\circ}-70^{\circ}-50^{\circ}=60^{\circ}$. The rest is analogous.

![](https://cdn.mathpix.com/cropped/2024_05_06_0ef0b80dacb6d98eff5ag-29.jpg?height=506&width=648&top_left_y=1233&top_left_x=705)

## Answer

$50^{\circ}, 60^{\circ}, 70^{\circ}$.