Answer
$A \approx70.5^\circ$
$B \approx70.5^\circ$
$C\approx 39.0^\circ$
Work Step by Step
1. Use the Law of Cosines, we have $A=cos^{-1}(\frac{(3)^2+(2)^2-(3)^2}{2(3)(2)})\approx70.5^\circ$
2. This is an isosceles triangle, we have $B=A\approx70.5^\circ$
3. We have $C\approx(180-70.5-70.5)=39.0^\circ$