Answer
$A \approx68.0^\circ$
$B\approx 44.0^\circ$
$C \approx68.0^\circ$
Work Step by Step
1. Use the Law of Cosines, we have $A=cos^{-1}(\frac{(3)^2+(4)^2-(4)^2}{2(3)(4)})\approx68.0^\circ$
2. It is an equilateral triangle, $B\approx(180-2(68.0))=44.0^\circ$
3. We have $C=A\approx68.0^\circ$
