Answer
$A =39.52^\circ$
$B =56.63^\circ$
$C=83.85^\circ$
Work Step by Step
$\cos{A} = \dfrac{b^2+c^2-a^2}{2bc} = \dfrac{(0.48)^2+(0.63)^2-(0.75)^2}{2 \times 0.48 \times 0.63} = 0.771$
$$\therefore A = \boxed{39.52^\circ}$$
$\cos{B} = \dfrac{a^2+c^2-b^2}{2ac} = \dfrac{(0.48)^2+(0.75)^2-(0.63)^2}{2 \times 0.48 \times 0.75} = 0.55$
$$\therefore B =\boxed{56.63^\circ} $$
$$C = 180-(A+B) = 180-(39.52+56.63) = \boxed{83.85^\circ}$$