Answer
$$c = 18.7\,{\text{cm}},{\text{ }}A = 91^\circ 40',\,\,B = 45^\circ 50'$$
Work Step by Step
$$\eqalign{
& a = {\text{27}}.{\text{6 cm}},b = {\text{19}}.{\text{8 cm}},C = {\text{42}}^\circ {\text{3}}0\prime \cr
& {\text{Use the Law of cosines to find }}c \cr
& {c^2} = {a^2} + {b^2} - 2ab\cos C \cr
& {\text{Substitute}} \cr
& {c^2} = {\left( {{\text{27}}.{\text{6}}} \right)^2} + {\left( {{\text{19}}.{\text{8}}} \right)^2} - 2\left( {{\text{27}}.{\text{6}}} \right)\left( {{\text{19}}.{\text{8}}} \right)\cos \left( {{\text{42}}^\circ {\text{3}}0\prime } \right) \cr
& {\text{Use a calculator}} \cr
& {c^2} = 347.985362 \cr
& {\text{Take square roots and choose the positive root}} \cr
& c \approx 18.7\,{\text{cm}} \cr
& \cr
& {\text{Calculate the angle }}A{\text{ using the law of sines}} \cr
& \frac{{\sin A}}{a} = \frac{{\sin C}}{c} \cr
& \sin A = \frac{{27.6\sin \left( {{\text{42}}^\circ {\text{3}}0\prime } \right)}}{{18.7}} \cr
& {\text{Use a calculator}} \cr
& \sin A = 0.9995712779 \cr
& A \approx {\sin ^{ - 1}}\left( {0.9995712779} \right) \cr
& A \approx 91^\circ 40' \cr
& \cr
& {\text{Calculate }}B \cr
& B = {180^ \circ } - A - C \cr
& B = {180^ \circ } - 91^\circ 40' - {\text{42}}^\circ {\text{3}}0\prime \cr
& B = 45^\circ 50' \cr
& \cr
& {\text{Answer}} \cr
& c = 18.7\,{\text{cm}},{\text{ }}A = 91^\circ 40',\,\,B = 45^\circ 50' \cr} $$
