## Task A-1.5.

Let $I$ be the center of the inscribed circle of an acute-angled triangle $ABC$ and let $|AC| > |BC|$. The angle bisector and the altitude from vertex $C$ intersect at an angle of $10^{\circ}$. If $\varangle AIB = 120^{\circ}$, determine the angles of triangle $ABC$.
## First Solution.

Let $\gamma = \angle ACB$. Let $N$ be the foot of the altitude from vertex $C$.

![](https://cdn.mathpix.com/cropped/2024_05_30_5bba13a74780d91c8bdbg-04.jpg?height=588&width=762&top_left_y=597&top_left_x=584)

According to the problem, $\angle BCN = \frac{\gamma}{2} - 10^\circ$.

Triangle $BCN$ is a right triangle, so $\angle CBN = 90^\circ - \left(\frac{\gamma}{2} - 10^\circ\right) = 100^\circ - \frac{\gamma}{2}$.

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In triangle $ABC$, $\angle CAB = 180^\circ - \gamma - \left(100^\circ - \frac{\gamma}{2}\right) = 80^\circ - \frac{\gamma}{2}$. 1 point

Since $I$ is the incenter, lines $AI$ and $BI$ are the angle bisectors of $\angle BAC$ and $\angle ACB$, respectively. Therefore, $\angle BAI = 40^\circ - \frac{\gamma}{4}$ and $\angle ABI = 50^\circ - \frac{\gamma}{4}$.

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From the sum of angles in triangle $ABI$, we have $40^\circ - \frac{\gamma}{4} + 50^\circ - \frac{\gamma}{4} + 120^\circ = 180^\circ$,

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which gives $\gamma = 60^\circ$.

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Finally, $\angle CBA = 100^\circ - \frac{\gamma}{2} = 70^\circ$ and $\angle BAC = 80^\circ - \frac{\gamma}{2} = 50^\circ$.

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