We have $\hat{BAD} = 180^\circ - (29^\circ + 41^\circ) = 110^\circ$, $\hat{BCD} = 82^\circ + 58^\circ = 140^\circ$. Consider the circumcircle $\gamma$ of $\triangle BCD$. Since $\hat{BAD} + \hat{BCD} > 180^\circ$, point $A$ is interior to $\gamma$.

Extend $CA$ beyond $A$ to meet $\gamma$ at $E$. By inscribed angles $\hat{EBD} = \hat{ECD} = \hat{ACD} = 58^\circ$, $\hat{EDB} = \hat{ECB} = \hat{ACB} = 82^\circ$.

Given that $\hat{ABD} = 29^\circ$, $\hat{ADB} = 41^\circ$ we obtain that $BA$ and $DA$ are bisectors of $\hat{EBD}$ and $\hat{EDB}$ respectively.

Hence $A$ is the incenter of triangle $BDE$, implying that $EA$ is the bisector of $\hat{BED}$.

From the cyclic quadrilateral $BCDE$ we have
$$\hat{BED} = 180^\circ - \hat{BCD} = \hat{BEC} = \frac{1}{2} \hat{BED} = 20^\circ \text{ and analogously}$$
$$\hat{DBC} = 20^\circ. \text{ In conclusion, } \hat{ABC} = 29^\circ + 20^\circ = 49^\circ, \hat{ADC} = 41^\circ + 20^\circ = 61^\circ.$$