Answer
$$A = {42^ \circ }00',\,\,\,\,B = {35^ \circ }50',\,\,\,C = {102^ \circ }10'$$
Work Step by Step
$$\eqalign{
& a = 42.9{\text{m,}}\,\,\,b = 37.6{\text{m,}}\,\,\,c = 62.7{\text{m}} \cr
& \cr
& {\text{Use the law of cosines to solve for any angle of the triangle}} \cr
& {c^2} = {a^2} + {b^2} - 2ab\cos C \cr
& {\text{Solve for cos }}C \cr
& \cos C = \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}} \cr
& {\text{Substitute}} \cr
& \cos C = \frac{{{{\left( {42.9} \right)}^2} + {{\left( {37.6} \right)}^2} - {{\left( {62.7} \right)}^2}}}{{2\left( {42.9} \right)\left( {37.6} \right)}} \cr
& {\text{Use a calculator}} \cr
& \cos C \approx - 0.2098894 \cr
& {\text{Use the inverse cosine function}} \cr
& C \approx {102^ \circ }10' \cr
& \cr
& {\text{Use the Law of sines to find the angle of }}B \cr
& \frac{{\sin B}}{b} = \frac{{\sin C}}{c} \cr
& \sin B = \frac{{b\sin C}}{c} \cr
& \sin B = \frac{{37.6\sin \left( {{{102}^ \circ }10'} \right)}}{{62.7}} \cr
& {\text{Use a calculator}} \cr
& \sin B \approx 0.58621138 \cr
& {\text{Use the inverse sine function}} \cr
& B \approx {35^ \circ }50' \cr
& \cr
& {\text{Calculate }}A \cr
& A = {180^ \circ } - B - C \cr
& A = {180^ \circ } - {35^ \circ }50' - {102^ \circ }10' \cr
& A = {42^ \circ }00' \cr
& \cr
& {\text{Answer}} \cr
& A = {42^ \circ }00',\,\,\,\,B = {35^ \circ }50',\,\,\,C = {102^ \circ }10' \cr} $$
