3.1. A tangent $FK$ is drawn to the circumcircle $K$ of triangle $FDC$, and $\angle KFC=58^{\circ}$. Points $K$ and $D$ lie on opposite sides of line $FC$, as shown in the figure. Find the acute angle between the angle bisectors of $\angle CFD$ and $\angle FCD$. Express your answer in degrees.

![](https://cdn.mathpix.com/cropped/2024_05_06_13964f40f25dd4c878ddg-1.jpg?height=759&width=831&top_left_y=1311&top_left_x=156)
Answer: 61

Solution. Let's refer to the picture:

![](https://cdn.mathpix.com/cropped/2024_05_06_13964f40f25dd4c878ddg-1.jpg?height=597&width=743&top_left_y=2214&top_left_x=151)
$\angle KFC$ is the angle between the chord and the tangent, which is equal to half the arc $FC$, just like the inscribed angle $\angle FDC$ that subtends it. Let $O$ be the point of intersection of the angle bisectors of $\triangle FCD$. Then $\angle FOC = 180^{\circ} - (\angle FCO + \angle CFO) = 180^{\circ} - \frac{1}{2} \cdot (\angle FCD + \angle CFD) = 180^{\circ} - \frac{1}{2} \cdot (180^{\circ} - \angle FDC) = 90^{\circ} + \frac{1}{2} \angle FDC = 90^{\circ} + 29^{\circ} = 119^{\circ}$. The required acute angle is $180^{\circ} - 119^{\circ} = 61^{\circ}$.

Remark. O is not the center of the circumscribed circle.