+33665 I take it C is the angle, and you are talking about a triangle in which case you can use the Cosine rule in the form
$$c^2=a^2+b^2-2abcos(C)$$ to find c.
$${\mathtt{c}} = {\sqrt{{{\mathtt{192}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{291}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{192}}{\mathtt{\,\times\,}}{\mathtt{291}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{41.3}}^\circ\right)}}} \Rightarrow {\mathtt{c}} = {\mathtt{193.896\: \!211\: \!065\: \!943\: \!792\: \!5}}$$
or c ≈193.9
I'll leave you to use the sine rule $$\frac{sin(A)}{a}=\frac{sin(C)}{c}$$ to find A, and similarly for B (or just use the fact that the three angles must add to 90° to get the last one).
+33665 I take it C is the angle, and you are talking about a triangle in which case you can use the Cosine rule in the form
$$c^2=a^2+b^2-2abcos(C)$$ to find c.
$${\mathtt{c}} = {\sqrt{{{\mathtt{192}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{291}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{192}}{\mathtt{\,\times\,}}{\mathtt{291}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{41.3}}^\circ\right)}}} \Rightarrow {\mathtt{c}} = {\mathtt{193.896\: \!211\: \!065\: \!943\: \!792\: \!5}}$$
or c ≈193.9
I'll leave you to use the sine rule $$\frac{sin(A)}{a}=\frac{sin(C)}{c}$$ to find A, and similarly for B (or just use the fact that the three angles must add to 90° to get the last one).
